Work, Energy, and Power

Need to look at peggy’s notes and add in more look at hers and match them more closely. Add in a bunch of stuff like simple motion and some of her early slides I like.

6.1 WorkWork is the exertion of a force in the direction that an object movesWork is tied to motion: No motion, no work!Equal to the magnitude of the force times the magnitude of the displacementUnits are Joules (1 J = 1 N-m)

For a force in line with the motion:

Ex: A weight lifter is bench pressing a barbell whose weight is 710N. He raises and lowers the barbell 0.65 m at a constant velocity. How much work is done raising the barbell? Lowering it?

The work lowering the weight is -461.5J

dFW 20 N Moves

20m

JmNdFW 5.46165.0710

If force is at an angle, only the portion of the force in the direction of the motion creates work:

Ex: How much work is done to pull a kid in a wagon for 22 meters with a force of 40N on a handle that is at a 48 degree angle from the ground?

dFW )cos(

θF

JmNdFW 835.58822)48cos(40)cos(

Force and direction of motion both matter in defining work! There is no work done by a force if it causes

no displacement Forces can do positive, negative, or zero

work. When a box is pushed on a flat floor for example

The normal force and gravity do not work, since they are perpendicular to the direction of motion

The person pushing the box does positive work, since she is pushing in the direction of motion.

Friction does negative work, since it points opposite the direction of motion

Cos (90º) = 0

Cos (0º) =+1

Cos (180º) =-1

Question: If a man holds a 50 kg box at armslength for 2 hours as he stands still, how much work does he do on the box?

Question: If a man holds a 50 kg box at arms length for 2 hours as he walks 1 km forward, how much work does he do on the box?

FBD

W =Fcos(θ)d

W =Fcos(θ)·0m = 0J

FBD

W =Fcos(θ)d

W =Fcos(90º)·1000m = 0J

Fapp

Fapp

mg

mg

cos(90º) = 0

Question: If a man lifts a 50 kg box 2.0 meters, how much work does he do on the box?

d = 2mFBD

Fapp

mg

W =Fcos(θ)d

W =mgcos(θ)d

W =50kg·9.8m/s2·cos(0º)·2m = 980J

Work Energy Theorem Work changes mechanical energy! Theorem relates work to this change. Deals only with the work done by the NET

FORCE, not individual forces. If an applied force does positive work on a

system, it tries to increase mechanical energy.

If an applied force does negative work, it tries to decrease mechanical energy.

The two forms of mechanical energy are called potential and kinetic energy.

v = instantaneous velocity (initial or final) – work is done to accelerate objects and change their velocity

Jane uses a vine wrapped around a pulley to lift a 70-kg Tarzan to a tree house 9.0 meters above the ground.

How much work does Jane do when she lifts Tarzan?

How much work does gravity do when Jane lifts Tarzan?

d = 9.0mFBD

Fapp

mg

Wj =Fcos(θ)d

Wj =mgcos(θ)d

Wnet = Wj+Wg = 0

Wj =70kg·9.8m/s2·cos(0º)·9m = +6174J

Wg = -6174J

Joe pushes a 10-kg box and slides it across the floor at constant velocity of 3.0 m/s. The coefficient of kinetic friction between the box and floor is 0.50.

How much work does Joe do if he pushes the box for 15 meters?

How much work does friction do as Joe pushes the box?

FBD

Fappfk

ΣF = 015m

ΣF = fk – Fapp = 0

fk = Fapp

W =Fapp·cos(θ)d

W =fk·cos(θ)d

W =μFN·cos(θ)d

W =0.5·10kg·9.8m/s2·cos(0º) ·15m=735J

W =μmg·cos(θ)d

Wnet = Wj + Wg = 0 Wg = -735J

A father pulls his child in a little red wagon with constant speed. If the father pulls with a force of 16 N for 10.0 m, and the handle of the wagon is inclined at an angle of 60º above the horizontal, how much work does the father do on the wagon?

10m

θ=60ºΣF = 0

W =Fapp·cos(θ)d

W =16N·cos(60º) ·10m = 80J

Kinetic energy Energy due to motion K = ½ m v2

K: Kinetic Energy m: mass in kg v: speed in m/s

Unit: Joules

How do we get to the kinetic from the equations we already know?

FdW

maF madFd

221 atd

)( 221 atmaFd

)( 2221 tamFd

atv

221 )(atmFd

221 mvKE

of KEKEW

221 mvFd

A 10.0 g bullet has a speed of 1.2 km/s. What is the kinetic energy of the bullet?

What is the bullet’s kinetic energy if the speed is halved?

What is the bullet’s kinetic energy if the speed is doubled?

K=1/2mv2m=10.0g =0.010kgv = 1.2km = 1200m

K=1/2·0.01Kg·(1200m/s)2= 7200J

K=1/2m(1/2v)2

K=1/2m1/4(v)2

K=1/4(1/2mv2)

K=1/4·7200J = 1800J

Ex: A 58 kg skier is coasting at 3.6 m/s down a 25o slope. The slope suddenly flattens out and he slows down to 1.2 m/s. How much work did the snow do on the skier?

Why did the angle not impact the work done?

Jsm

smkgmvmvKEKEW ff 08.334))6.3()2.1((585.02121 222

02

0

6.3 Work done by GravityGravitational Potential Energy – energy due to an objects position relative to the earthThe object has the potential to do work if it can fall because of gravity

Units: Joules (J)

6.4 Conservation of Mechanical EnergyTotal mechanical energy (kinetic + potential) of an object remains constant provided the net work done is by gravity

hgmPEWork gravity

ffoo PEKEPEKE

madFdW

Ex:One of the fastest roller coasters in the world is the Magnum at Cedar Point. It includes a drop of 59.4 meters. Neglecting friction and starting with a speed at the top of 0 m/s, how fast is the coaster going at the bottom of the hill?

Equation from previous section:

020

2 2/12/1 mghmvmghmv ff

ffoo PEKEPEKE

smmmsmsmhhgvv ff /121.34)04.59(8.92)0()(2 220

20

axvv 220

2

Ex: A motorcycle is trying to leap across a canyon by driving off the higher cliff at 38 m/s. Find the impact velocity at which the cycle strikes the ground at the top of the far side of the cliff.

70m

35m

38 m/s

020

2 2/12/1 mghmvmghmv ff

2 2 20 02 ( ) (0 ) 2 9.8 (70 35 ) 26.160 /f fv v g h h m s m s m m m s

38m/s

26.16m/s

a2 + b2 = c2

c2=√( a2 + b2)

c2=√( (38m/s)2 + (26.16m/s)2) = 46.13m/s

θ

θ = tan-1(O/A)

A=O tan θ

θ = tan-1((26.16m/s)/(38m/s)) = 34.54º

Work Energy Theorem Result of doing work is a change in kinetic

energy. Theorem relates work to this change. Deals only with the work done by the NET

FORCE, not individual forces. The net work due to all forces equals the

change in the kinetic energy of a system.

Wnet = ΔK Wnet: work due to all forces acting on an

object ΔK: change in kinetic energy (Kf – Ki)

W1+W2+W3+ …..Wn = ΔK

A 15-g acorn falls from a tree and lands on the ground 10.0 m below with a speed of 11.0 m/s.

What would the speed of the acorn have been if there had been no air resistance and determine if 11.0m/s is real world velocity or not?

Did air resistance do positive, negative or zero work on the acorn? Why?

W =Fapp·cos(θ)d

Fapp = mg

W =0.015kg·9.8m/s2cos(0º) · 10m = 1.47J

mg d, define down direction as 0º

KE=0.5mv2

v =√ (KE/0.5m)= √ (1.47J/(0.5·0.015kg))= 14m/s

v2f =v2

i +2ad

vf =√((0m/s)+2 · -9.8m/s2 · -10m)= 14 m/s

d, define down direction as 0º

Fr

W =Fr·cos(180º)d = -J

W =Fapp·cos(θ)d

A 15-g acorn falls from a tree and lands on the ground 10.0 m below with a speed of 11.0 m/s.

How much work was done by air resistance?

What was the average force of air resistance?

Ki = 0

Wnet = Wg + Wd

ΔK = Kf - Ki

Wg + Wd = Kf-Ki

Wnet = ΔK Wg + Wd = Kf

Wd = 1/2·0.015kg·(11 m/s)2- 1.47J

Wd = 1/2mv2- Wg

Wd = -0.5625J

Wd =Fd·cos(θ)d

Fd=Wd/(cos(θ)d)

Fd=-0.5625J / cos(180º)·10m

Fd= 0.05625 N

6.7 Power Rate at which work is done by a force.Rate at which energy is changing.

Net force can sometimes be found using Newton’s 2nd law, F=ma.If the object is moving up or down with constant speed, the force is the weight.

W F dPower

t t

SI unit for Power is the Watt. 1 Watt = 1 Joule/s Named after the Scottish engineer James Watt (1776-

1819) who perfected the steam engine. British system

horsepower 1 hp = 746 W

The kilowatt-hour is a commonly used unit by the electrical power company. Power companies charge you by the kilowatt-hour

(kWh), but this not power, it is really energy consumed. 1 kW = 1000 W 1 h = 3600 s 1 kWh = 1000J/s • 3600s = 3.6 x 106J

Ex: A 1100kg car starts from rest and accelerates for 5 seconds at 4.6m/s/s. How much power does the car generate?

5060 57.558190 /

5

W F d N mPower J s

t t s

NsmkgmaF 5060/6.41100 2

mssmatd 5.57)5(/6.45.2/1 222

A record was set for stair climbing when a man ran up the 1600 steps of the Empire State Building in 10 minutes and 59 seconds. If the height gain of each step was 0.20 m, and the man’s mass was 70.0 kg, what was his average power output during the climb? Give your answer in both watts and horsepower.

H of 1 step is 0.2 m

h = 0.2m · 1600 steps = 320 m

W =Fapp·cos(θ)d

W =mg·cos(θ)d

Fapp =mg

W =70kg·9.8m/s2·cos(0º) · 320m = 219520 J

t = 10min · 60s = 600s + 59s = 659s

P= 219520J / 659s = 333.111 W P= W/t

Calculate the power output of a 1.0 g fly as it walks straight up a window pane at 2.5 cm/s. The window pane is 0.4 meters tall.

v = d/t

P= 3.13 x 10-7 J / 16s = 1.956x10-8 W

P= W/t

m = 1.0g = 0.001kg

K=1/2 · 0.001 kg (0.025m/s)2 = 3.13 x 10 -7 J

K=1/2mv2

v = 2.5cm/s = 1 m / 100cm = 0.025m/s

KE= W = 3.13 x 10-7 J

t=v/d

t=0.4 m / 0.025m/s = 16s

Constant force and work The force shown is a constant force. W = F·Δd can be used to calculate the work done by this force when it moves an object from xa to xb. The area under the curve from xa to xb can also be used to calculate the work done by the force when it moves an object from xa to xb

F(x)

x

xa xb

Δd

F

Area of a square = l·w

W = l·w =F·Δd = area under the curve

The force shown is avariable force. W = F·Δd CANNOT be used to calculate thework done by thisforce! The area under thecurve from xa to xb canSTILL be used tocalculate the workdone by the forcewhen it moves anobject from xa to xb xa xb

Area of a square = l·w

Area of a triangle = 0.5bh

W = l·w + 0.5bh =F·Δd = area under the curve

F(x)

x

When a spring is stretched or compressed from its equilibrium position, it does negative work, since the spring pulls opposite the direction of motion. Ws = - ½ k x2

Ws: work done by spring (J) k: force constant of spring (N/m) x: displacement from equilibrium (m)

The force doing the stretching does positive work equal to the magnitude of the work done by the spring.

Wapp = - Ws = ½ k x2

0

M

M

x

Fs

200

-200

100

-100

0

F(N)

x(m)

10 2 3 4 5

Ws = negative area = - ½ kx2

Fs = -kx (Hooke’s Law)

Cont with slide 25