Work, power energy Q&A

8/12/2019 Work, power energy Q&A

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Ch6. Work and Energy

Work Done by a Constant Force

Force Fpoints in the same direction as the

resulting displacement s

W=Fs


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SI Unit of Work: newton.meter=joule(J)

System Force * Distance = Work

SI newton(N) meter(m) joule(J)

CGS dyne(dyn) centimeter(cm) erg

BE pound(lb) foot(ft) foot.pound(ft.lb)

sFW )cos(

Units of Measurement for Work


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Example 1. Pulling a Suitcase-

on-Wheels


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Find the work done by a 45.0 N force in

pulling the suitcase at an angle fora distance s=75.0 m 0.50

JmNsFW 2170)0.75(0.50cos)0.45()cos(

No work done due to Fsin

F and s in the same direction positive work

F and s in the opposite direction negative work


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Example 2. Bench-Pressing

The weight lifter is bench-pressing a barbell whoseweight is 710N. In part (b) of the figure, he raises

the barbell a distance of 0.65m above his chest, and

in part (c) he lowers it the same distance.


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The weight is raised and lowered at a constant

velocity. Determine the work done on thebarbell by the weight lifter during (a) the

lifting phase and (b) the lowering phase.

(a)

(b)

JmNsFW 460)65.0(0cos)710()cos(

JmNsFW 460)65.0(180cos)710()cos(

cos1800 = -1


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Example 3. Accelerating a Crate

A 120kg crate on the flatbed of a

truck that is moving with an

acceleration of a=+1.5m/s2along

the positive x axis. The crate does

not slip with respect to the truck,

as the truck undergoes a

displacement whose magnitude iss=65m. What is the total work

done on the crate by all of the

forces acting on it?


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Forces that act on the crate:

(1) the weight W=mg of the crate,

(2) the normal force FNexerted by the flatbed,

(3) the static frictional force fs.

Nsmkgmafs 180)/5.1)(120( 2

JmNsfW s4102.1)65(0cos)180()cos(


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Check your understanding 1A suitcase is hanging straight down from your hand as you

ride an escalator. Your hand exerts a force on the suitcase

and this force does work.Which one of the following

statements is correct?

(a) The work is negative when you ride up the escalator and

positive when you ride down the escalator.

(b) The work is positive when you ride up the escalator and

negative when you ride down the escalator.

(c) The work is positive irrespective of whether you ride up or

down the escalator.

(d) The work is negative irrespective of whether you ride up

or down the escalator.

Answer: (b)


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The Work-Energy Theorem and

Kinetic Energy


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Concepts At a Glance:

In physics, when a net force performs work on anobject, there is always a result from the effort.

The result is a change in the Kinetic energy.

What is Kinetic Energy?

Energy associated with motion.

KE=(1/2)mv2

The relationship that relates work to the change

in kinetic energy is known as the Work-energy

theorem.


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Definition of Kinetic Energy:

The kinetic energy KE of an object with mass m

and speed vis given by

2

2

1mvKE

SI Unit of Kinetic Energy: joule(J)

asvvf 2

2

0

2 +a

vv

s

f

2

2

0

2


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massF)(Work done by net ext. force

2

0

2

2

1

2

1)( mvmvsF f Work done by

net ext. forceFinal KE Initial KE


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The Work-Energy Theorem

When a net external force does work W on

an object, the kinetic energy of the object

changes from its initial value of KE0to a final

value of KEf, the difference between the twovalues being equal to the work:

20

20

21

21 mvmvKEKEW ff


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Example 4. Deep Space 1


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The space probe Deep Space 1 was launched October 24,

1998. Its mass was 474kg. The goal of the mission was totest a new kind of engine called an ion propulsion drive,

which generates only a weak thrust, but can do so for

long periods of time using only small amounts of fuel. The

mission has been spectacularly successful. Consider theprobe traveling at an initial speed of v0=275m/s. No forces

act on it except the 56.0-mN thrust of its engine. This

external force F is directed parallel to the displacement s

of magnitude 2.42*109m. Determine the final speed of the

probe, assuming that the mass remains nearly constant.


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)1042.2(0cos)100.56()cos( 93

mNsFW

=1.36*108J

28

0 )/275)(474(

2

1)1036.1( smkgJKEWKEf ++

=1.54*108J

mKEv ff )(2 kg

J474

)1054.1(2 8

=806 m/s


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Example 5. Downhill Skiing


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A 58 kg skier is coasting down a 250slope. A kinetic

frictional force of magnitude fk=70N opposes hermotion. Near the top of the slope, the skiers speed

is v0=3.6m/s. Ignoring air resistance, determine the

speed vfat a point that is displaced 57m downhill.

NsmkgfmgF k 7025sin)/80.9)(58(25sin 2

=+170 N


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KEf=W+KE0=9700J+(1/2)(58kg)(3.6m/s)2=10100J

m

KEv

f

f

)(2 sm

kg

J/19

58

)10100(2

JmNsFW 9700)57(0cos)170()cos(


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Check your understanding 2

A rocket is at rest on the launch pad. When the

rocket is launched, its kinetic energy increases.

Is the following statement true or false?

The amount by which the kinetic energy

increases is equal to the work done by the force

generated by the rockets engine.

Answer: False


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Conceptual Example 6. Work and

Kinetic EnergyA satellite moving abut the earth in a

circular orbit and in an elliptical

orbit. The only external force that acts

on the satellite is the gravitational

force. For these two orbits, determine

whether the kinetic energy of the

satellite changes during the motion.

KE changes in the elliptical orbit,

but not in the circular orbit.

In circular motion, F S alwaysno work done.


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Gravitational Potential Energy

Work Done by the Force of Gravity

Wgravity=(mg cos00)(h0-hf)=mg(h0-hf)


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Example 7. A Gymnast on a

Trampoline


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A gymnast springs vertically upward from a

trampoline. The gymnast leaves the trampoline at a

height of 1.20m and reaches a maximum height of

4.80m before falling back down. All heights are

measured with respect to the ground. Ignoring air

resistance, determine the initial speed v0with whichthe gymnast leaves the trampoline.

)80.420.1(/80.9(2)(2 200 mmsmhhgv f

=8.40 m/s

Wgravity=mg (h0-hf)


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Gravitational Potential Energy

Wgravity= mgh0 - mghf

Initial

gravitationa

l potential

energy PE0

Final

gravitational

potential

energy PEf


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Definition of Gravitational Potential Energy

The gravitational potential energy PE is the energy

that an object of mass m has by virtue of its

position relative to the surface of the earth. That

position is measured by the height h of the objectrelative to an arbitrary zero level:

PE=mgh

SI Unit of gravitational potential energy: joule (J)


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Conservative Versus Non-

conservative Forces


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Definition of a Conservative Force

Version 1: A force is conservative when the work it

does on a moving object is independent of the path

between the objects initial and final positions.

Version 2: A force is conservative when it does no

net work on an object moving around a closed path,

starting and finishing at the same point.


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20

20

21

21)( mvmvWhhmg fncf +

Work done by external forces =

KE

i.e,2

0

2

2

1

2

1

mvmvWW fncc +


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PEKEWnc +

Wnc = (KEf - KE0) + (PEf - PE0)Change in

kinetic energy

Change in gravitational

potential energy

Net work

done by non-

conservative

forces

)()2

1

2

1

( 02

0

2

mghmghmvmvW ffnc +


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The Conservation of Mechanical

Energy


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SupposeWnc=0J, so Ef = E0

mvf2+mghf= mv0

2+mgh0

The principle of Conservation of Mechanical Energy

The total mechanical energy (E=KE+PE) of an

object remains constant as the object moves,

provided that the net work done by external non-conservative forces is zero, Wnc=0J

2

1

2

1


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Example 8. A Daredevil

Motorcyclist


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A motorcyclist is trying to leap across the canyon by

driving horizontally off the cliff at a speed of 38.0m/s. Ignoring air resistance, find the speed with

which the cycle strikes the ground on the other side.

(1/2)mvf2

+mghf= (1/2) mv02

+mgh0

)(2 02

0 ff hhgvv +

)0.350.70)(/80.9(2)/9.38( 22 mmsmsmvf +

=46.2 m/s


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Check your understanding 3

Some of the following situations are consistent with the principleof conservation of mechanical energy, and some are not.

Which ones are consistent with the principle?

(a) An object moves uphill with an increasing speed.

(b) An object moves uphill with a decreasing speed.

(c) An object moves uphill with a constant speed.

(d) An object moves downhill with an increasing speed.

(e) An object moves downhill with a decreasing speed.

(f) An object moves downhill with a constant speed.

(b) And (d)


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Conceptual Example 9. The

Favorite Swimming HoleA rope is tied to a tree limb

and used by a swimmer to

swing into the water below.The person starts from rest

with the rope held in the

horizontal position, swings

downward, and then lets goof the rope.


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Three forces act on him: his weight, the tension in the rope, and

the force due to air resistance. His initial height h0and final

height hf are known. Considering the nature of these forces,

conservative versus non-conservative, can we use the principle

of conservation of mechanical energy to find his speed vfat the

point where he lets go of the rope?

If Wnc= 0 (work due to nonconservative forces)

conservation of energy

Tension and air resistance --- non conservative

So no work done by T.

T is always r to the circular path.


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Work done by air resistance is nonzero.

So ideally no.

But if ignore air resistance,

ff mghmvmghmv ++ 2

0

2

02

1

2

1

)(2 02

0 ff hhgvv +


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Example 10. The Steel Dragon

The tallest and fastest roller coaster in the world is now

the Steel Dragon in Mie, Japan. The ride includes a

vertical drop of 93.5m. The coaster has a speed of

3.0m/s at the top of the drop. Neglect friction and findthe speed of the riders at the bottom.

(1/2)mvf2+mghf= (1/2) mv0

2+mgh0

Ef E0


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)(2 02

0 ff hhgvv +

)5.93)(/80.9(2)/0.3( 2 msmsmvf +

= 42.9m/s (about 96 mi/h)


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Example 11. The Steel Dragon,

RevisitedIn example 10, we ignored non-conservative forces,

such as friction. In reality, however, such forces are

present when the roller coaster descends. The actualspeed of the riders at the bottom is 41.0 m/s, which is

less than that determined in example 10. Assuming

again that the coaster has a speed of 3.0 m/s at the top,

find the work done by non-conservative forces on a55.0kg rider during the descent from a height h0to a

height hf, where h0hf=93.5m


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)

2

1()

2

1( 0

2

0

2mghmvmghmvW ffnc ++

)()(2

10

2

0

2

ffnc hhmgvvmW

EfE0

)5.93)(/80.9)(0.55(])/3()/41)[(0.55(

2

1 222 msmkgsmsmkgWnc

J4400


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Example 12. FireworksA 0.20kg rocket in a fireworks

display is launched from rest and

follows an erratic flight path to

reach the point P. Point P is 29m

above the starting point. In theprocess, 425J of work is done on

the rocket by the non-conservative

force generated by the burning

propellant. Ignoring air resistanceand the mass lost due to the

burning propellant, find the speed

vfof the rocket at the point P.


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)2

1()

2

1( 0

2

0

2mghmvmghmvW ffnc ++

m

hhmgmvW

vfnc

f

)](2

1[2

0

2

0 +

kg

msmkgsmkgJ

vf20.0

)]29)(/80.9)(20.0()/0)(20.0(2

1425[2

22 +

=61m/s


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Power

The idea of power incorporates both the concepts ofwork and time, for poweris work done per unit time.

Definition of Average Power

Average power P is the average rate at which work Wis done, and it is obtained by dividing W by the time t

required to perform the work.

tW

TimeWorkP

SI Unit of Power: joule/s=watt (W)


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Check your understanding 4Engine A has a greater power rating than engine B. Which one

of the following statements correctly describes the abilities

of these engines to do work?

(a) Engine A and B can do the same amount of work in the

same amount of time.

(b) In the same amount of time, engine B can do more work

than engine A.

(c) Engine A and B can do the same amount of work, but A cando it more quickly.

Answer: (c)


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Units of Measurement for Power

System Work / Time = Power

SI joule (J) / second (s) = watt (W)

CGS erg / second (s) = erg persecond(erg/s)

BE foot.pound / second (s) = foot.pound

per second

(ft.lb) (ft.lb/s)


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Time

energyinchangeP

wattsondpoundfoothorsepower 7.745sec/5501

t

Fs

t

W

vFP


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smssmsmatvvf /0.23)00.5)(/60.4()/0( 2

0 ++

)0.78(1082.5)/5.11)(5060( 4 hpWsmNvFP

ff vvvv 2

1)(2

10 +


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Other Forms of Energy and the

Conservation of EnergyExamples: electrical energy, heat, chemical energy,

nuclear energy.

In general, energy of all types can be converted

from one form to another.

The principle of conservation of energy

Energy can neither be created nor destroyed, but can

only be converted from one form to another.


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Work Done by a Variable ForceThe work done by a variable force in moving an

object is equal to the area under the graph of Fcos

versus s.


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++ 2211 )cos()cos( sFsFW


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Example 14. Work and the

Compound BowFind the work that the archer must do in drawing backthe string of the compound bow from 0 to 0.500 m

Jsquare

J

squaresW 5.60)250.0)(242(


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Example 15. Skateboarding and

Work

The skateboarder is coasting down a ramp, and there

are three forces acting on her: her weight W(magnitude = 675 N), a frictional force f ( magnitude =

125 N). That opposes her motion, and a normal force

FN (magnitude =612 N). Determine the net work done

by the three forces when she coasts for a distance of9.2 m.


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Force F Angle S W= (Fcos sW 675N 65.00 9.2m W=(675N)(cos 65.00)(9.2m)=+2620J

f 125N 180.00 9.2m W=(125N)(cos 180.00)(9.2m)=-1150J

FN 612N 90.00 9.2m W=(612N)(cos 90.00)(9.2m)=0J

The net work done by the three forces is

+2626J+(-1150J)+0J=+1470J


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Concepts & Calculations Example 16.

Conservation of Mechanical Energy and

the Work-Energy Theorem

hA


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A 0.41kg block sliding from A to B along a frictionless

surface. When the block reaches B, it continues to

slide along the horizontal surface BC. The blockslows down, coming to rest at C. The kinetic energy

of the block at A is 37J, and the height of A and B

are 12m and 7m above the ground.

(a) What is the kinetic energy of the block when itreaches B?

(b) How much work does the kinetic frictional force do

during the BC segment of the trip?


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A---B motion: What are the forces?

Weight---------------

What is the non conservative work (Wnc) ?

Wnc=0 FN displacement

Conservation of energy valid for A---B?

K. E at B ? K. E at A

K. E at B < K. E at A

Normal force--------

conservative

non conservative


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(a) KEB + mghB = KEA + mghA

KEB = KEA + mg(hA-hB)

=37J+(0.41kg)(9.80m/s2)(12m-7m)

=57J


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Problem 2

REASONING AND SOLUTION Each locomotive does work

W= T(cos s= (5.00103N) cos 20.0 (2.00 103m) = 9.40106JThe net work is then

WT= 2W= 2T(cos s = 1.88 107J

T

T


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Problem 16

2820 m/s

apogee8450 m/s

perigee


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From the work-energy theorem,

1 1 12 2 2 2f 0 f 02 2 2

W mv mv m v v

1 2 2

2W (7420 kg) (2820 m/s) (8450 m/s) =

a)

b) J111035.2


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Problem 20

To find the work, employ the work-energy theorem,

W= KEf KE0

F = 24N

8.0 M

v0= 0 vf= 2.0 m/8

16 kg

FN

P

mg

fkfk= kFN FN-mg=0

FN=mg


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SOLUTION According to the work-energy theorem, we have

W = Wpull+ Wf= KEf KE0

Using Equation 6.1 [W= (Fcos ) s] to express each work

contribution, writing the kinetic energy as , and

noting that the initial kinetic energy is zero (the sled starts

from rest), we obtain

212mv

) ) 212kpull f

cos0 cos180P s f s mv

W W

+

Work done by the net force (pulling force P and fk)

W= Wpulling

+ Wf


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Solving for the coefficient of kinetic friction gives

) )

) ) ) )

) ) )

221 12 2

k 2

cos0 16 kg 2.0 m/s 24 N 8.0 m0.13

cos180 16 kg 9.80 m/s 8.0 m

mv P s

mg s

The angle between the force and the displacement is 0 for

the pulling force (it points in the same direction as the

displacement) and 180 for the frictional force (it points

opposite to the displacement). Equation 4.8 indicates that the

magnitude of the frictional force is fk = kFN, and we know

that the magnitude of the normal force is FN= mg. With these

substitutions the work-energy theorem becomes

) ) 212kpull f

cos0 cos180P s mg s mv

W W

+


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Problem 26

REASONING The work done by the weight of

the basketball is given by Equation 6.1

as , where F= mgis the

magnitude of the weight, is the angle

between the weight and the displacement, and

sis the magnitude of the displacement. The

drawing shows that the weight and

displacement are parallel, so that = 0.Thepotential energy of the basketball is given byEquation 6.5 as PE = mgh, where his the

height of the ball above the ground.

)cosW F ss

mg

6.1 m

1.5 m

=0.6g


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SOLUTION

a. The work done by the weight of the basketball is

)cosW F s = mg (cos 0)(h0hf)= (0.60 kg)(9.80 m/s2)(6.1 m 1.5 m) = 27 J

b. The potential energy of the ball, relative to the ground,

when it is released is

PE0= mgh0= (0.60 kg)(9.80 m/s2)(6.1 m) = 36 J


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d. The change in the balls gravitational potential energy is

PE = PEfPE0= 8.8 J36 J = 27 J

We see that the change in the gravitational potential

energy is equal to27 J = , where Wis the work

done by the weight of the ball (see part a).W

c. The potential energy of the ball, relative to the ground,

when it is caught is

PEf= mghf= (0.60 kg)(9.80 m/s2)(1.5 m) = 8.8 J


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Problem 34

Total mechanical energy

conserved?

2 21 1

f f 0 02 2

mv mgh mv mgh+ +

) )

)

2 2

f 0 2

14.0 m/s 13.0 m/s1.4 m

2 9.80 m/sh h

Yes (WconservativeFN displacement)


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Problem 41

REASONING Friction and

air resistance are being

ignored. The normal force

from the slide is

perpendicular to the motion,

so it does no work. Thus, no

net work is done by non-

conservative forces, and the

principle of conservation ofmechanical energy applies.


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SOLUTION Applying the principle of conservation of

mechanical energy to the swimmer at the top and the bottom

of the slide, we have

1

2mvf

2 +mghf

Ef

1

2mv0

2 +mgh0

E0

If we let hbe the height of the bottom of the slide above the

water, , and . Since the swimmer starts from

rest, m/s, and the above expression becomes

12vf2 +ghgH

v0 0hf h h0 H


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The vertical displacement of the swimmer after

leaving the slide is, from Equation 3.5b (with down

being negative),

Therefore, h= 1.23 m. Using these values of and h

in the above expression forH, we find

Hh+ vf2

2g 1.23 m + (10.0 m/s)

2

2(9.80 m/s2

) 6.33 m

vf

) ) mssmtay y 23.1500.0/80.92

1

2

1 222


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REASONING AND SOLUTION At the bottom of the

circular path of the swing, the centripetal force isprovided by the tension in the rope less the weight of the

swing and rider. That is,

C

2

F

mvT mg

r

Solving for the mass yields

2Tm

vg

r

+


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The energy of the swing is conserved if friction is ignored.

The initial energy,E0, when the swing is released is

completely potential energy and isE0= mgh0,

Conservation of energy

PEini+ KEini= PEf+ KEf

h0= r(1cos 60.0) =12r

r=2h0

mgh0+ 0 = 0 + (1/2)mvf2

0= 2v gh gr f


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The expression for the mass now becomes

)

2

2

8.00 10 N40.8 kg

2 2 9.80 m/s

Tm

g

gr

ghT

gr

vTm

+

+

02 2

g

T

gg

T

2

+


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Problem 48

m=0.75kg

18.0 m/s

No air friction.


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a. Since there is no air friction, the only force that acts on the

projectile is the conservative gravitational force (its weight). The

initial and final speeds of the ball are known, so the

conservation of mechanical energy can be used to find the

maximum height that the projectile attains.

The conservation of mechanical energy, as expressed by

Equation 6.9b, states that

2 21 1f f 0 02 2

0f

mv mgh mv mgh

E E

+ +


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The mass mcan be eliminated algebraically from this

equation since it appears as a factor in every term. Solving

for the final height hf

gives

)2 21 0 f2f 0

v vh h

g

+

Setting h0= 0 m and vf= 0 m/s, the final height, in the

absence of air resistance, is

) )

)2 22 2

o ff 2

18.0 m/s 0 m/s16.5 m

2 2 9.80 m/ s

v vh

g


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The work-energy theorem is

) )2 21 1nc f 0 f 02 2W mv mv mgh mgh +

b. When air resistance, a non-conservative force, is present, it

does negative work on the projectile and slows it down.

Consequently, the projectile does not rise as high as when thereis no air resistance. The work-energy theorem, in the form of

Equation 6.6, may be used to find the work done by air friction.

Then, using the definition of work, Equation 6.1, the average

force due to air resistance can be found.


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where Wncis the non-conservative work done by air

resistance. According to Equation 6.1, the work can be

written as , where is the average

force of air resistance. As the projectile moves upward, the

force of air resistance is directed downward, so the angle

between the two vectors is = 180and cos =1. Themagnitude sof the displacement is the difference betweenthe final and initial heights, s= hfh0= 11.8 m. With these

substitutions, the work-energy theorem becomes

sFW Rnc )180cos(

) )2 21

R f o f 02F s m v v mg h h +

RF


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Solving for givesRF

) )

) ) ) ) ) ) )

2 21f o f 02

R

2 2 212

0.750 kg 0 m/s 18.0 m/s 0.750 kg 9.80 m/s 11.8 m2.9 N

11.8 m

m v v mg h hF

s

+

+


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Problem 57

REASONING AND SOLUTION One is the amount of

work or energy generated when one kilowatt of power

is supplied for a time of one hour. From

Equation 6.10a, we know that . Using the fact

that and that 1h = 3600 s, we have

W P t1 kW = 1.0 103J/s

1.0 kWh = (1.0 103J/s)(1 h) = (1.0103 J/s)(3600 s) = 3.6106J

bl


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Problem 60

REASONING AND SOLUTION

a. The power developed by the engine is

P= Fv= (2.00 102N)(20.0 m/s) = 4.00 103W

mg

37

friction


93/93

The power developed by the engine is then

P= Fv= (Fa+ mgsin 37.0)v

P= [2.00 102N + (2.50 102kg)(9.80 m/s2)sin 37.0](20.0 m/s)

3.35 10 4 W=

b. The force required of the engine in order to

maintain a constant speed up the slope is

F= Fa+ mg sin 37.0

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Work, power energy Q&A