cb

aopp

osit

e

adja

cen

t to

adjacent to opposite

Two angles whose sum is 90deg are called

complementary angles

hypotenuseopposite

opposite

c

b

hypotenuse

oppositesin

adjacentcos

hypotenuse

a

c

oppositetan

adjacent

b

a

c

a

hypotenuse

oppositesin

c

b

hypotenuse

adjacentcos

b

a

adjacent

oppositetan

Shrink yourself down and stand where the angle is

2

c

c c

aa

a

bb

b

3

4

What is the system

a small portion of the Universe

A valid system is

bull may be a single object or particlebull may be a collection of objects or particlesbull may be a region of space (such as the interior of an automobile engine combustioncylinder)bull may vary with time in size and shape (such as a rubber ball which deformsupon striking a wall)

a system boundary an imaginary surface (not necessarily coinciding with a physical surface) that divides the Universe into the system and the environment surroundingthe system

5

bull W = F Δx cos bull Work = the product of force and displacement

times the cosine of the angle between the force and the direction of the displacement

bull How much work is done pulling with a 15 N force applied at 20o over a distance of 12 m

bull W = FΔx cos bull W = 15 N (12 m) (cos 20) = 169J

20o

15 N

W = F Δx units Nm or joule j

bullScalar quantitybullIndependent of time

ve work Force acts in the opposite direction as the displacement

object loses energy

Sign of work

cos 90 = 0

No work is done on the load

When force F is at right angles to displacement s (F s) perp

F Δ Δxx cos = 0

No work is done ifFF = 0 orΔΔx x = 0 or = 90o

6

+ve work Force acts in the same direction as the displacement

object gains energy

Example When Work is Zerobull A man carries a bucket of water horizontally at constant

velocitybull The force does no work on the bucketbull Displacement is horizontalbull Force is verticalbull cos 90deg = 0

7

example

8

examplebull An Eskimo returning pulls a sled as shown The total mass of

the sled is 500 kg and he exerts a force of 120 times 102 N on the sled by pulling on the rope How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

J

mN

xFW

2

2

1025

)05)(30)(cos10201(

)cos(

9

bull Suppose microk = 0200 How much work done on the sled by friction and the net work if θ = 30deg and he pulls the sled 50 m

sin

0sin

FmgN

FmgNF ynet

J

mN

smkg

xFmgxN

xfxfW

kk

kkfric

2

2

2

1034

)05)(30sin1021

89050)(2000(

)sin(

)180cos(

J

JJ

WWWWW gNfricFnet

090

0010341025 22

Example A horizontal force F pulls a 10 kg carton across the floor

at constant speed If the coefficient of sliding friction between the carton and the floor is 030 how much work

is done by F in moving the carton by 5m

The carton moves with constant speed Thus the carton is in horizontal equilibrium F = f = μk N = μk mg Thus F = 03 x 10 x 98= 294 N Therefore work done W = FS=(294 Cos 0o)=147 J

10

11

7-4 Work Done by a Varying Force

12

13

Same units as work

Remember the Eq of motion

Multiply both sides by m 1

2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi

2

2ax

KE 1

2mv2

When work is done on a system and the only change in the system is in its speed the work done by the net force equals the change in kinetic energy of the system

The speed of the system increases if the work done on it is positive ( the kinetic energy of the particle increases) The speed of the system decreases if the net work done on it is negative( the kinetic energy of the object decreases)

W = F xW = K Work-Energy Theorem

KE f KE i Wnet

14

An objectrsquos kinetic energy depends on its mass and its speed

The faster something is moving and the heavier it is the more work it can do so hellip

If lsquovrsquo quadruples

١٤٤٤ ١٠ ١ 15

then KE will increase by 9 timesIf lsquovrsquo tripled

then KE will increase by 16 times

M= 4 kg

example

bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy

16

bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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• Slide 52

2

c

c c

aa

a

bb

b

3

4

What is the system

a small portion of the Universe

A valid system is

bull may be a single object or particlebull may be a collection of objects or particlesbull may be a region of space (such as the interior of an automobile engine combustioncylinder)bull may vary with time in size and shape (such as a rubber ball which deformsupon striking a wall)

a system boundary an imaginary surface (not necessarily coinciding with a physical surface) that divides the Universe into the system and the environment surroundingthe system

5

bull W = F Δx cos bull Work = the product of force and displacement

times the cosine of the angle between the force and the direction of the displacement

bull How much work is done pulling with a 15 N force applied at 20o over a distance of 12 m

bull W = FΔx cos bull W = 15 N (12 m) (cos 20) = 169J

20o

15 N

W = F Δx units Nm or joule j

bullScalar quantitybullIndependent of time

ve work Force acts in the opposite direction as the displacement

object loses energy

Sign of work

cos 90 = 0

No work is done on the load

When force F is at right angles to displacement s (F s) perp

F Δ Δxx cos = 0

No work is done ifFF = 0 orΔΔx x = 0 or = 90o

6

+ve work Force acts in the same direction as the displacement

object gains energy

Example When Work is Zerobull A man carries a bucket of water horizontally at constant

velocitybull The force does no work on the bucketbull Displacement is horizontalbull Force is verticalbull cos 90deg = 0

7

example

8

examplebull An Eskimo returning pulls a sled as shown The total mass of

the sled is 500 kg and he exerts a force of 120 times 102 N on the sled by pulling on the rope How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

J

mN

xFW

2

2

1025

)05)(30)(cos10201(

)cos(

9

bull Suppose microk = 0200 How much work done on the sled by friction and the net work if θ = 30deg and he pulls the sled 50 m

sin

0sin

FmgN

FmgNF ynet

J

mN

smkg

xFmgxN

xfxfW

kk

kkfric

2

2

2

1034

)05)(30sin1021

89050)(2000(

)sin(

)180cos(

J

JJ

WWWWW gNfricFnet

090

0010341025 22

Example A horizontal force F pulls a 10 kg carton across the floor

at constant speed If the coefficient of sliding friction between the carton and the floor is 030 how much work

is done by F in moving the carton by 5m

The carton moves with constant speed Thus the carton is in horizontal equilibrium F = f = μk N = μk mg Thus F = 03 x 10 x 98= 294 N Therefore work done W = FS=(294 Cos 0o)=147 J

10

11

7-4 Work Done by a Varying Force

12

13

Same units as work

Remember the Eq of motion

Multiply both sides by m 1

2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi

2

2ax

KE 1

2mv2

When work is done on a system and the only change in the system is in its speed the work done by the net force equals the change in kinetic energy of the system

The speed of the system increases if the work done on it is positive ( the kinetic energy of the particle increases) The speed of the system decreases if the net work done on it is negative( the kinetic energy of the object decreases)

W = F xW = K Work-Energy Theorem

KE f KE i Wnet

14

An objectrsquos kinetic energy depends on its mass and its speed

The faster something is moving and the heavier it is the more work it can do so hellip

If lsquovrsquo quadruples

١٤٤٤ ١٠ ١ 15

then KE will increase by 9 timesIf lsquovrsquo tripled

then KE will increase by 16 times

M= 4 kg

example

bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy

16

bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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3

4

What is the system

a small portion of the Universe

A valid system is

bull may be a single object or particlebull may be a collection of objects or particlesbull may be a region of space (such as the interior of an automobile engine combustioncylinder)bull may vary with time in size and shape (such as a rubber ball which deformsupon striking a wall)

a system boundary an imaginary surface (not necessarily coinciding with a physical surface) that divides the Universe into the system and the environment surroundingthe system

5

bull W = F Δx cos bull Work = the product of force and displacement

times the cosine of the angle between the force and the direction of the displacement

bull How much work is done pulling with a 15 N force applied at 20o over a distance of 12 m

bull W = FΔx cos bull W = 15 N (12 m) (cos 20) = 169J

20o

15 N

W = F Δx units Nm or joule j

bullScalar quantitybullIndependent of time

ve work Force acts in the opposite direction as the displacement

object loses energy

Sign of work

cos 90 = 0

No work is done on the load

When force F is at right angles to displacement s (F s) perp

F Δ Δxx cos = 0

No work is done ifFF = 0 orΔΔx x = 0 or = 90o

6

+ve work Force acts in the same direction as the displacement

object gains energy

Example When Work is Zerobull A man carries a bucket of water horizontally at constant

velocitybull The force does no work on the bucketbull Displacement is horizontalbull Force is verticalbull cos 90deg = 0

7

example

8

examplebull An Eskimo returning pulls a sled as shown The total mass of

the sled is 500 kg and he exerts a force of 120 times 102 N on the sled by pulling on the rope How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

J

mN

xFW

2

2

1025

)05)(30)(cos10201(

)cos(

9

bull Suppose microk = 0200 How much work done on the sled by friction and the net work if θ = 30deg and he pulls the sled 50 m

sin

0sin

FmgN

FmgNF ynet

J

mN

smkg

xFmgxN

xfxfW

kk

kkfric

2

2

2

1034

)05)(30sin1021

89050)(2000(

)sin(

)180cos(

J

JJ

WWWWW gNfricFnet

090

0010341025 22

Example A horizontal force F pulls a 10 kg carton across the floor

at constant speed If the coefficient of sliding friction between the carton and the floor is 030 how much work

is done by F in moving the carton by 5m

The carton moves with constant speed Thus the carton is in horizontal equilibrium F = f = μk N = μk mg Thus F = 03 x 10 x 98= 294 N Therefore work done W = FS=(294 Cos 0o)=147 J

10

11

7-4 Work Done by a Varying Force

12

13

Same units as work

Remember the Eq of motion

Multiply both sides by m 1

2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi

2

2ax

KE 1

2mv2

When work is done on a system and the only change in the system is in its speed the work done by the net force equals the change in kinetic energy of the system

The speed of the system increases if the work done on it is positive ( the kinetic energy of the particle increases) The speed of the system decreases if the net work done on it is negative( the kinetic energy of the object decreases)

W = F xW = K Work-Energy Theorem

KE f KE i Wnet

14

An objectrsquos kinetic energy depends on its mass and its speed

The faster something is moving and the heavier it is the more work it can do so hellip

If lsquovrsquo quadruples

١٤٤٤ ١٠ ١ 15

then KE will increase by 9 timesIf lsquovrsquo tripled

then KE will increase by 16 times

M= 4 kg

example

bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy

16

bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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4

What is the system

a small portion of the Universe

A valid system is

bull may be a single object or particlebull may be a collection of objects or particlesbull may be a region of space (such as the interior of an automobile engine combustioncylinder)bull may vary with time in size and shape (such as a rubber ball which deformsupon striking a wall)

a system boundary an imaginary surface (not necessarily coinciding with a physical surface) that divides the Universe into the system and the environment surroundingthe system

5

bull W = F Δx cos bull Work = the product of force and displacement

times the cosine of the angle between the force and the direction of the displacement

bull How much work is done pulling with a 15 N force applied at 20o over a distance of 12 m

bull W = FΔx cos bull W = 15 N (12 m) (cos 20) = 169J

20o

15 N

W = F Δx units Nm or joule j

bullScalar quantitybullIndependent of time

ve work Force acts in the opposite direction as the displacement

object loses energy

Sign of work

cos 90 = 0

No work is done on the load

When force F is at right angles to displacement s (F s) perp

F Δ Δxx cos = 0

No work is done ifFF = 0 orΔΔx x = 0 or = 90o

6

+ve work Force acts in the same direction as the displacement

object gains energy

Example When Work is Zerobull A man carries a bucket of water horizontally at constant

velocitybull The force does no work on the bucketbull Displacement is horizontalbull Force is verticalbull cos 90deg = 0

7

example

8

examplebull An Eskimo returning pulls a sled as shown The total mass of

the sled is 500 kg and he exerts a force of 120 times 102 N on the sled by pulling on the rope How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

J

mN

xFW

2

2

1025

)05)(30)(cos10201(

)cos(

9

bull Suppose microk = 0200 How much work done on the sled by friction and the net work if θ = 30deg and he pulls the sled 50 m

sin

0sin

FmgN

FmgNF ynet

J

mN

smkg

xFmgxN

xfxfW

kk

kkfric

2

2

2

1034

)05)(30sin1021

89050)(2000(

)sin(

)180cos(

J

JJ

WWWWW gNfricFnet

090

0010341025 22

Example A horizontal force F pulls a 10 kg carton across the floor

at constant speed If the coefficient of sliding friction between the carton and the floor is 030 how much work

is done by F in moving the carton by 5m

The carton moves with constant speed Thus the carton is in horizontal equilibrium F = f = μk N = μk mg Thus F = 03 x 10 x 98= 294 N Therefore work done W = FS=(294 Cos 0o)=147 J

10

11

7-4 Work Done by a Varying Force

12

13

Same units as work

Remember the Eq of motion

Multiply both sides by m 1

2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi

2

2ax

KE 1

2mv2

When work is done on a system and the only change in the system is in its speed the work done by the net force equals the change in kinetic energy of the system

The speed of the system increases if the work done on it is positive ( the kinetic energy of the particle increases) The speed of the system decreases if the net work done on it is negative( the kinetic energy of the object decreases)

W = F xW = K Work-Energy Theorem

KE f KE i Wnet

14

An objectrsquos kinetic energy depends on its mass and its speed

The faster something is moving and the heavier it is the more work it can do so hellip

If lsquovrsquo quadruples

١٤٤٤ ١٠ ١ 15

then KE will increase by 9 timesIf lsquovrsquo tripled

then KE will increase by 16 times

M= 4 kg

example

bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy

16

bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

• Slide 1
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• Slide 52

5

bull W = F Δx cos bull Work = the product of force and displacement

times the cosine of the angle between the force and the direction of the displacement

bull How much work is done pulling with a 15 N force applied at 20o over a distance of 12 m

bull W = FΔx cos bull W = 15 N (12 m) (cos 20) = 169J

20o

15 N

W = F Δx units Nm or joule j

bullScalar quantitybullIndependent of time

ve work Force acts in the opposite direction as the displacement

object loses energy

Sign of work

cos 90 = 0

No work is done on the load

When force F is at right angles to displacement s (F s) perp

F Δ Δxx cos = 0

No work is done ifFF = 0 orΔΔx x = 0 or = 90o

6

+ve work Force acts in the same direction as the displacement

object gains energy

Example When Work is Zerobull A man carries a bucket of water horizontally at constant

velocitybull The force does no work on the bucketbull Displacement is horizontalbull Force is verticalbull cos 90deg = 0

7

example

8

examplebull An Eskimo returning pulls a sled as shown The total mass of

the sled is 500 kg and he exerts a force of 120 times 102 N on the sled by pulling on the rope How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

J

mN

xFW

2

2

1025

)05)(30)(cos10201(

)cos(

9

bull Suppose microk = 0200 How much work done on the sled by friction and the net work if θ = 30deg and he pulls the sled 50 m

sin

0sin

FmgN

FmgNF ynet

J

mN

smkg

xFmgxN

xfxfW

kk

kkfric

2

2

2

1034

)05)(30sin1021

89050)(2000(

)sin(

)180cos(

J

JJ

WWWWW gNfricFnet

090

0010341025 22

Example A horizontal force F pulls a 10 kg carton across the floor

at constant speed If the coefficient of sliding friction between the carton and the floor is 030 how much work

is done by F in moving the carton by 5m

The carton moves with constant speed Thus the carton is in horizontal equilibrium F = f = μk N = μk mg Thus F = 03 x 10 x 98= 294 N Therefore work done W = FS=(294 Cos 0o)=147 J

10

11

7-4 Work Done by a Varying Force

12

13

Same units as work

Remember the Eq of motion

Multiply both sides by m 1

2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi

2

2ax

KE 1

2mv2

When work is done on a system and the only change in the system is in its speed the work done by the net force equals the change in kinetic energy of the system

The speed of the system increases if the work done on it is positive ( the kinetic energy of the particle increases) The speed of the system decreases if the net work done on it is negative( the kinetic energy of the object decreases)

W = F xW = K Work-Energy Theorem

KE f KE i Wnet

14

An objectrsquos kinetic energy depends on its mass and its speed

The faster something is moving and the heavier it is the more work it can do so hellip

If lsquovrsquo quadruples

١٤٤٤ ١٠ ١ 15

then KE will increase by 9 timesIf lsquovrsquo tripled

then KE will increase by 16 times

M= 4 kg

example

bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy

16

bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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ve work Force acts in the opposite direction as the displacement

object loses energy

Sign of work

cos 90 = 0

No work is done on the load

When force F is at right angles to displacement s (F s) perp

F Δ Δxx cos = 0

No work is done ifFF = 0 orΔΔx x = 0 or = 90o

6

+ve work Force acts in the same direction as the displacement

object gains energy

Example When Work is Zerobull A man carries a bucket of water horizontally at constant

velocitybull The force does no work on the bucketbull Displacement is horizontalbull Force is verticalbull cos 90deg = 0

7

example

8

examplebull An Eskimo returning pulls a sled as shown The total mass of

the sled is 500 kg and he exerts a force of 120 times 102 N on the sled by pulling on the rope How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

J

mN

xFW

2

2

1025

)05)(30)(cos10201(

)cos(

9

bull Suppose microk = 0200 How much work done on the sled by friction and the net work if θ = 30deg and he pulls the sled 50 m

sin

0sin

FmgN

FmgNF ynet

J

mN

smkg

xFmgxN

xfxfW

kk

kkfric

2

2

2

1034

)05)(30sin1021

89050)(2000(

)sin(

)180cos(

J

JJ

WWWWW gNfricFnet

090

0010341025 22

Example A horizontal force F pulls a 10 kg carton across the floor

at constant speed If the coefficient of sliding friction between the carton and the floor is 030 how much work

is done by F in moving the carton by 5m

The carton moves with constant speed Thus the carton is in horizontal equilibrium F = f = μk N = μk mg Thus F = 03 x 10 x 98= 294 N Therefore work done W = FS=(294 Cos 0o)=147 J

10

11

7-4 Work Done by a Varying Force

12

13

Same units as work

Remember the Eq of motion

Multiply both sides by m 1

2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi

2

2ax

KE 1

2mv2

When work is done on a system and the only change in the system is in its speed the work done by the net force equals the change in kinetic energy of the system

The speed of the system increases if the work done on it is positive ( the kinetic energy of the particle increases) The speed of the system decreases if the net work done on it is negative( the kinetic energy of the object decreases)

W = F xW = K Work-Energy Theorem

KE f KE i Wnet

14

An objectrsquos kinetic energy depends on its mass and its speed

The faster something is moving and the heavier it is the more work it can do so hellip

If lsquovrsquo quadruples

١٤٤٤ ١٠ ١ 15

then KE will increase by 9 timesIf lsquovrsquo tripled

then KE will increase by 16 times

M= 4 kg

example

bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy

16

bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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Example When Work is Zerobull A man carries a bucket of water horizontally at constant

velocitybull The force does no work on the bucketbull Displacement is horizontalbull Force is verticalbull cos 90deg = 0

7

example

8

examplebull An Eskimo returning pulls a sled as shown The total mass of

the sled is 500 kg and he exerts a force of 120 times 102 N on the sled by pulling on the rope How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

J

mN

xFW

2

2

1025

)05)(30)(cos10201(

)cos(

9

bull Suppose microk = 0200 How much work done on the sled by friction and the net work if θ = 30deg and he pulls the sled 50 m

sin

0sin

FmgN

FmgNF ynet

J

mN

smkg

xFmgxN

xfxfW

kk

kkfric

2

2

2

1034

)05)(30sin1021

89050)(2000(

)sin(

)180cos(

J

JJ

WWWWW gNfricFnet

090

0010341025 22

Example A horizontal force F pulls a 10 kg carton across the floor

at constant speed If the coefficient of sliding friction between the carton and the floor is 030 how much work

is done by F in moving the carton by 5m

The carton moves with constant speed Thus the carton is in horizontal equilibrium F = f = μk N = μk mg Thus F = 03 x 10 x 98= 294 N Therefore work done W = FS=(294 Cos 0o)=147 J

10

11

7-4 Work Done by a Varying Force

12

13

Same units as work

Remember the Eq of motion

Multiply both sides by m 1

2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi

2

2ax

KE 1

2mv2

When work is done on a system and the only change in the system is in its speed the work done by the net force equals the change in kinetic energy of the system

The speed of the system increases if the work done on it is positive ( the kinetic energy of the particle increases) The speed of the system decreases if the net work done on it is negative( the kinetic energy of the object decreases)

W = F xW = K Work-Energy Theorem

KE f KE i Wnet

14

An objectrsquos kinetic energy depends on its mass and its speed

The faster something is moving and the heavier it is the more work it can do so hellip

If lsquovrsquo quadruples

١٤٤٤ ١٠ ١ 15

then KE will increase by 9 timesIf lsquovrsquo tripled

then KE will increase by 16 times

M= 4 kg

example

bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy

16

bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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example

8

examplebull An Eskimo returning pulls a sled as shown The total mass of

the sled is 500 kg and he exerts a force of 120 times 102 N on the sled by pulling on the rope How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

J

mN

xFW

2

2

1025

)05)(30)(cos10201(

)cos(

9

bull Suppose microk = 0200 How much work done on the sled by friction and the net work if θ = 30deg and he pulls the sled 50 m

sin

0sin

FmgN

FmgNF ynet

J

mN

smkg

xFmgxN

xfxfW

kk

kkfric

2

2

2

1034

)05)(30sin1021

89050)(2000(

)sin(

)180cos(

J

JJ

WWWWW gNfricFnet

090

0010341025 22

Example A horizontal force F pulls a 10 kg carton across the floor

at constant speed If the coefficient of sliding friction between the carton and the floor is 030 how much work

is done by F in moving the carton by 5m

The carton moves with constant speed Thus the carton is in horizontal equilibrium F = f = μk N = μk mg Thus F = 03 x 10 x 98= 294 N Therefore work done W = FS=(294 Cos 0o)=147 J

10

11

7-4 Work Done by a Varying Force

12

13

Same units as work

Remember the Eq of motion

Multiply both sides by m 1

2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi

2

2ax

KE 1

2mv2

When work is done on a system and the only change in the system is in its speed the work done by the net force equals the change in kinetic energy of the system

The speed of the system increases if the work done on it is positive ( the kinetic energy of the particle increases) The speed of the system decreases if the net work done on it is negative( the kinetic energy of the object decreases)

W = F xW = K Work-Energy Theorem

KE f KE i Wnet

14

An objectrsquos kinetic energy depends on its mass and its speed

The faster something is moving and the heavier it is the more work it can do so hellip

If lsquovrsquo quadruples

١٤٤٤ ١٠ ١ 15

then KE will increase by 9 timesIf lsquovrsquo tripled

then KE will increase by 16 times

M= 4 kg

example

bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy

16

bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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examplebull An Eskimo returning pulls a sled as shown The total mass of

the sled is 500 kg and he exerts a force of 120 times 102 N on the sled by pulling on the rope How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

J

mN

xFW

2

2

1025

)05)(30)(cos10201(

)cos(

9

bull Suppose microk = 0200 How much work done on the sled by friction and the net work if θ = 30deg and he pulls the sled 50 m

sin

0sin

FmgN

FmgNF ynet

J

mN

smkg

xFmgxN

xfxfW

kk

kkfric

2

2

2

1034

)05)(30sin1021

89050)(2000(

)sin(

)180cos(

J

JJ

WWWWW gNfricFnet

090

0010341025 22

Example A horizontal force F pulls a 10 kg carton across the floor

at constant speed If the coefficient of sliding friction between the carton and the floor is 030 how much work

is done by F in moving the carton by 5m

The carton moves with constant speed Thus the carton is in horizontal equilibrium F = f = μk N = μk mg Thus F = 03 x 10 x 98= 294 N Therefore work done W = FS=(294 Cos 0o)=147 J

10

11

7-4 Work Done by a Varying Force

12

13

Same units as work

Remember the Eq of motion

Multiply both sides by m 1

2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi

2

2ax

KE 1

2mv2

When work is done on a system and the only change in the system is in its speed the work done by the net force equals the change in kinetic energy of the system

The speed of the system increases if the work done on it is positive ( the kinetic energy of the particle increases) The speed of the system decreases if the net work done on it is negative( the kinetic energy of the object decreases)

W = F xW = K Work-Energy Theorem

KE f KE i Wnet

14

An objectrsquos kinetic energy depends on its mass and its speed

The faster something is moving and the heavier it is the more work it can do so hellip

If lsquovrsquo quadruples

١٤٤٤ ١٠ ١ 15

then KE will increase by 9 timesIf lsquovrsquo tripled

then KE will increase by 16 times

M= 4 kg

example

bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy

16

bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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Example A horizontal force F pulls a 10 kg carton across the floor

at constant speed If the coefficient of sliding friction between the carton and the floor is 030 how much work

is done by F in moving the carton by 5m

The carton moves with constant speed Thus the carton is in horizontal equilibrium F = f = μk N = μk mg Thus F = 03 x 10 x 98= 294 N Therefore work done W = FS=(294 Cos 0o)=147 J

10

11

7-4 Work Done by a Varying Force

12

13

Same units as work

Remember the Eq of motion

Multiply both sides by m 1

2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi

2

2ax

KE 1

2mv2

When work is done on a system and the only change in the system is in its speed the work done by the net force equals the change in kinetic energy of the system

The speed of the system increases if the work done on it is positive ( the kinetic energy of the particle increases) The speed of the system decreases if the net work done on it is negative( the kinetic energy of the object decreases)

W = F xW = K Work-Energy Theorem

KE f KE i Wnet

14

An objectrsquos kinetic energy depends on its mass and its speed

The faster something is moving and the heavier it is the more work it can do so hellip

If lsquovrsquo quadruples

١٤٤٤ ١٠ ١ 15

then KE will increase by 9 timesIf lsquovrsquo tripled

then KE will increase by 16 times

M= 4 kg

example

bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy

16

bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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11

7-4 Work Done by a Varying Force

12

13

Same units as work

Remember the Eq of motion

Multiply both sides by m 1

2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi

2

2ax

KE 1

2mv2

When work is done on a system and the only change in the system is in its speed the work done by the net force equals the change in kinetic energy of the system

The speed of the system increases if the work done on it is positive ( the kinetic energy of the particle increases) The speed of the system decreases if the net work done on it is negative( the kinetic energy of the object decreases)

W = F xW = K Work-Energy Theorem

KE f KE i Wnet

14

An objectrsquos kinetic energy depends on its mass and its speed

The faster something is moving and the heavier it is the more work it can do so hellip

If lsquovrsquo quadruples

١٤٤٤ ١٠ ١ 15

then KE will increase by 9 timesIf lsquovrsquo tripled

then KE will increase by 16 times

M= 4 kg

example

bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy

16

bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

• Slide 1
• Slide 2
• Slide 3
• Slide 4
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• Slide 49
• Slide 50
• Slide 51
• Slide 52

12

13

Same units as work

Remember the Eq of motion

Multiply both sides by m 1

2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi

2

2ax

KE 1

2mv2

When work is done on a system and the only change in the system is in its speed the work done by the net force equals the change in kinetic energy of the system

The speed of the system increases if the work done on it is positive ( the kinetic energy of the particle increases) The speed of the system decreases if the net work done on it is negative( the kinetic energy of the object decreases)

W = F xW = K Work-Energy Theorem

KE f KE i Wnet

14

An objectrsquos kinetic energy depends on its mass and its speed

The faster something is moving and the heavier it is the more work it can do so hellip

If lsquovrsquo quadruples

١٤٤٤ ١٠ ١ 15

then KE will increase by 9 timesIf lsquovrsquo tripled

then KE will increase by 16 times

M= 4 kg

example

bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy

16

bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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13

Same units as work

Remember the Eq of motion

Multiply both sides by m 1

2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi

2

2ax

KE 1

2mv2

When work is done on a system and the only change in the system is in its speed the work done by the net force equals the change in kinetic energy of the system

The speed of the system increases if the work done on it is positive ( the kinetic energy of the particle increases) The speed of the system decreases if the net work done on it is negative( the kinetic energy of the object decreases)

W = F xW = K Work-Energy Theorem

KE f KE i Wnet

14

An objectrsquos kinetic energy depends on its mass and its speed

The faster something is moving and the heavier it is the more work it can do so hellip

If lsquovrsquo quadruples

١٤٤٤ ١٠ ١ 15

then KE will increase by 9 timesIf lsquovrsquo tripled

then KE will increase by 16 times

M= 4 kg

example

bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy

16

bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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• Slide 2
• Slide 3
• Slide 4
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• Slide 52

Same units as work

Remember the Eq of motion

Multiply both sides by m 1

2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi

2

2ax

KE 1

2mv2

When work is done on a system and the only change in the system is in its speed the work done by the net force equals the change in kinetic energy of the system

The speed of the system increases if the work done on it is positive ( the kinetic energy of the particle increases) The speed of the system decreases if the net work done on it is negative( the kinetic energy of the object decreases)

W = F xW = K Work-Energy Theorem

KE f KE i Wnet

14

An objectrsquos kinetic energy depends on its mass and its speed

The faster something is moving and the heavier it is the more work it can do so hellip

If lsquovrsquo quadruples

١٤٤٤ ١٠ ١ 15

then KE will increase by 9 timesIf lsquovrsquo tripled

then KE will increase by 16 times

M= 4 kg

example

bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy

16

bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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An objectrsquos kinetic energy depends on its mass and its speed

The faster something is moving and the heavier it is the more work it can do so hellip

If lsquovrsquo quadruples

١٤٤٤ ١٠ ١ 15

then KE will increase by 9 timesIf lsquovrsquo tripled

then KE will increase by 16 times

M= 4 kg

example

bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy

16

bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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• Slide 2
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bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy

16

bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example

17

18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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18

Kinetic Energy ndash the energy of motionbull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joulebull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2 K = frac12 (6 kg) (4 ms)sup2 KE = 48 JKE = frac12 mt vtsup2 v = radic2 Km = 620 ms

19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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19

M=600 kg

Another solution

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

smxav

xavv

smmaF

4632

2

26

12

m

Fa

21

022

2

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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• Slide 2
• Slide 3
• Slide 4
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• Slide 52

Example A 60-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N Find the speed of the block after it has moved 30 m W =F d cos 0 =12x 3x1 = 36J Δk = w 05m V2 =W =36 J

Jv 36)(62

1 2

smv 46312

20

Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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Example What acceleration is required to stop a 1000kg car traveling 28 ms in a distance of 100 meters

2

2

2

923200

784

100)28(2

12

1

sma

a

xmavm

21

22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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• Slide 2
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22

example

example

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

• Slide 1
• Slide 2
• Slide 3
• Slide 4
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• Slide 49
• Slide 50
• Slide 51
• Slide 52

exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m

24

bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work

25

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
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• Slide 47
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• Slide 51
• Slide 52

Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 26

Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU

27

the work Wint done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value

28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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28

Elastic Potential Energy

The elastic potential energy function associated with the blockndashspring system is

Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm = 0200 kg x 98 Nkg-1 x 030 m = 059 J

change in gpe = 059 J

example

042123 29Norah Ali Al Moneef king Saud university

What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m

example

Potential Energy = weight x height changeWeight = 12 x 10 = 120 NHeight change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105 J

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example

30

١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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١٤٤٤ ١٠ ١ 31Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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١٤٤٤ ١٠ ١ 32Norah Ali Al - moneef

Who is going faster at the bottombull Assume no frictionbull Assume both have the same

speed pushing off at the top

same

example

7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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7-7Conservative and Nonconservative Forces

bull Law of conservation of energy- energy cannot be created or destroyed

closed system- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the

system may not be at present at the end

33

Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh34

bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example

042123 35

exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1

042123 36

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example

042123 37

B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms

042123 38

example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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example

At B

At C

39

A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2mFind the upward speed of truck after its displacement

40

Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00

41

Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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Work and Energybull If a force (other than gravity) acts on the system and does workbull Need Work-Energy relation

bull orffii KEPEWKEPE

22

2

1

2

1ffii mvmghWmvmgh

42

On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))

Δx = 247 m

example

042123 43Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0

042123 44Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example

042123 45Norah Ali Al Moneef king Saud university

ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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ExampleSuppose the initial kinetic and potential energies of a system are 75J and 250J respectively and that the final kinetic and potential energies of the same system are 300J and -25J respectively How much work was done on the system by non-conservative forces 1 0J 2 50J 3 -50J 4 225J 5 -225J

correct

Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies

W = (300-75) + ((-25) - 250) = 225 - 275 = -50J

١٤٤٤ ١٠ ١ 46Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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١٤٤٤ ١٠ ١ 47Norah Ali Al - moneef

exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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exampleYou are towing a car up a hill with constant velocity

A )The work done on the car by the normal force is

1 positive2 negative3 zero

W

T

FN V

The normal force is perpendicular to the displacement hence does no work

correct

b )The work done on the car by the gravitational force is

1 positive2 negative3 zero

correctWith the surface defined as the x-axis the x component of gravity is in the opposite direction of the displacement therefore work is negative

C ) The work done on the car by the tension force is

1 positive2 negative3 zero

correct

Tension is in the same direction as the displacement

١٤٤٤ ١٠ ١ 48Norah Ali Al - moneef

١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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١٤٤٤ ١٠ ١ 49Norah Ali Al - moneef

50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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50

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =

042123 Norah Ali Al Moneef king Saud university

In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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In a rifle barrel a 150 g bullet is accelerated from rest to a speed of 780 ms

a)Find the work that is done on the bullet

example

Using Work-Kinetic Energy Theorem W = ΔKE W = KEf - KEi W = 12(0015 kg)(780 ms)2 - 0 W = 456 kJ

b) If the rifle barrel is 720 cm long find the magnitude of the average total force that acted on the bullet

Using W = Fcosθd F = (456 kJ) (072 m) F = 633 kN

١٤٤٤ ١٠ ١ 51Norah Ali Al - moneef

Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

042123 52Norah Ali Al Moneef king Saud university

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