17/8/04

Midterm 1: July 9

Will cover material from Chapters 1-6 Go to the room where you usually have

recitation Practice exam available on-line and in the

library

7/8/04 2

Chapter 7

Kinetic Energy and Work

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Energy

A property of the state of an object Scalar quantity – no direction Conserved – cannot be created or

destroyed, but it can change from one form to another or be exchanged from one object to another

Units: Joule = kg m2/s2

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Kinds of Energy

Kinetic (movement) Potential

GravitationalSpringChemical bonds

Mass (E=mc2…) Thermal And more…

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Kinetic Energy

Kinetic Energy Energy of motion

For an object of mass m moving with speed v:

Also kinetic energy associated with rotation, vibration, etc.

221 mvK

More on that later…

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Kinetic Energy: Orders of Magnitude

Earth orbiting sun: 2x1029 J

Car at 60 mph: 100,000 J

Baseball pitch: 300 J

K = ½mv2 for some common objects:

A man walking: 40 J

Angry bee: 0.005 J

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Work

W > 0 energy added W < 0 energy taken away

Work Energy transferred by a force

Work done on an object is the energy transferred to/from it

Work done on an object by a constant force F while moving through a displacement r

W = F • r

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is the angle between the vectors if you put their tails together

Dot or Scalar Product

zzyyxx BABABABABA cos

B

A

Measures “how much” one vector lies along another

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What Does It Mean Physically?

W > 0 if < 90° force is adding energyto object

W < 0 if > 90° force is reducing energy of object

W = 0 if F r

r

F

r

FrF

FrrF

||

cos

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Work Examples

Ftot = 0 Wtot = 0 Note: Kinetic Energy is constant…

Push on a wallr = 0, so no work is done (W = 0)

d

Lift a weight against gravity at constant speed…

mgddFW

mgddFW

liftlift

gravgrav

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Work Against Gravity

A weightlifter does work when lifting a weight

Wlift=mgh

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Work for Tension, Gravity, Friction

What is the work done by gravity, tension, and friction?

Iguazu Falls: 269 ft = 82 m

Guess that the weight of the pack is 250 lb. = 113 kg

Vertical ascent: Wgravity = -mgh = -(113 kg)(9.8 m/s2)(82 m) = -90,800 J

Wtension = -Wgravity = 90,800 J

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Horizontal pull (at the end): use d~50 ft=15.2 m, μk = 1.0

Wfriction = Ff d = -μkmg d = -(1.0)(113 kg)(9.8 m/s2)(15.2 m) = -16,800 JWtension = -Wfriction = 16,800 J

What is the work done by gravity, tension, and friction?

Work for Tension, Gravity, Friction

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Work Due to Friction

vFfr

The frictional force always opposes the motion:

Moving to the right: Moving to the left:

W negative in both cases

0

x

xFW fr

0

x

xFW fr

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Another Work ExampleConsider a pig sliding down frictionless ramp:

N

Fg

d

θ

dh

0 dNdNWNormal

mgh

mgd

mgd

dFWgrav

sin

)90cos(

Work done by the normal force:

Work from the gravitational force:

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Work-Kinetic Energy Theorem

The total work done on an object is the change in its kinetic energy!

Easy to extrapolate to three dimensions

For constant acceleration:

Since Ftot = ma:

Multiply by ½m:

But ½mv2 = K: xFKK xtotif ,

)(222ifif xxavv

)(2 ,22

ifxtot

if xxm

Fvv

)(,2

212

21

ifxtotif xxFmvmv

totWdFK

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Total Work in One Dimension

For a small segment x,W F(x)x

From x1 to x2:

Wtot F(x)x = area under curvex1 < x < x2

To be exact: 2

1

)(x

xdxxFW

x1 x2

x

F

x

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Work in Two Dimensions

In one dimension motion and force are always in the same direction (or opposite directions)

This is not true in two dimensions

F

dr

How do we generalize work and kinetic energy to motion in two dimensions?

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Work Along an Arbitrary Path

F

Δx

Δyl

Over the whole path:

2

1

2

1

2

1

z

z z

y

y y

x

x x

finish

start

dzFdyFdxF

ldFW

lF

yFxFW yx

Looking at a small patch of the path:

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Restating the Work-Kinetic Energy Theorem

dFmv

)( 221

Constant force:

ldFmv

)( 221

Variable force:

WK

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A Formal Derivation

f

i

x

x

F(x) dxW = f

i

x

x

dxa(x)m=W

dx

dv=v

dt

dx

dx

dv=

dt

dva(x)=

f

i

f

i

x

x

v

v

dvvdx=mdx

dvvW=m

22

2

2

1

2

1

2

1

if

v

v

v

v

mvmv=

mv=dvvm=Wf

i

f

i

Newton’s 2nd Law

Using the chain rule:

227/8/04

An Example

vi=60 mph =26.8 m/s

150 ftμk=0.9

45.7 m

x

Will the car be able to stop before hitting the moose?

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An Example (continued)

Friction from the tires on the road will slow the car:

mgNF kkfr

The work done by friction will be:

xmg

xFW

k

frfr

From the work-energy theorem:

xmgmvmv

WKK

kif

if

2

212

21

Just miss the moose!

mx

sm

sm

g

vx

k

f

8.40

)/8.9)(9.0(

)/8.26(

2

1

2

12

22

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Example: Pile Driver

d

Big Mass

Drop a big mass to drive a nail into a board

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The Pile Driver

h

x

mHow much force is exerted on the nail?

1) Work done by gravity on freely falling pile

W1 = (-mg)(-h-x) = mg(h+x)

2) Work done on the nail

W2 = F(-x) = -Fx

3) Total Work = 0

W = W1+W2 = mg(h+x) – Fx = 0

F=mg(h+x)

x

267/8/04

Example: Slowing a Bus

J. =

m/s). kg) (.( =

mv = K ii

6

2321

221

1003

53310455

J x . =

m/s). kg) ( x .( =

m v=K ff

6

2321

221

1032

02910455

vi=75 mph=33.5 m/s

How much work is done slowing down a bus?

5.45 x 103 kg

vf=65 mph=29.0 m/s

JW

JJKKW if

5

66

107

100.3103.2

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Example: (continued)

Fbrake = 2.75 x 104 N

How far does the bus travel while slowing?

f

i

x

x

F(x) dxW =

)-x (xW = F ifbrake

Area under F(x) curve:x

F

xi xf

Fbrake

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)( ifbrake -xx W = Fbrake

if F

W -xx )(

24

225

4

5

10752

107

10752

107)(

kg m/s.

/s kg m

N.-

J--xx if

m.-xx if 25)(

23

4

0.51045.5

10752m/s

kg

N.

m

Fa brake

Example: (continued)

297/8/04

The Spring: A Variable Force

Springs exert force when stretched or compressed

positive x

x = 0 Defines equilibrium position

F = - kx

k = "spring constant"big k stiff spring

x = 0 no force x < 0 F > 0 spring pushes outx > 0 F < 0 spring pulls in

Hooke’s Law:

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Work Done By a Spring

How much work by spring in moving from xi to xf?

xf

If |xi| < |xf| then spring takes energy away

xi x=0

Fspring Fspring

x=0

22

2

1

2

1)()( fi

x

x

x

xkxkxdxkxdxxFW

f

i

f

i

317/8/04

Example:

You stretch a spring 10 cm and must apply a 10 N force to hold the spring in place. What is the spring constant, and how much work did you do on the spring to stretch it?

mNm

N

x

Fk

kxF

spring

spring

/1001.0

)10(

10 cm

FpullFspring

327/8/04

Example:10 cm

FpullFspring

The work done by the spring is:

J

mmN

kx

kxkxW

f

fispring

5.0

)1.0)(/100( 221

2

21

2

212

21

So the work I do on the spring is

JWW springpull 5.0

337/8/04

PowerWork doesn’t depend on the time interval

Work to climb a flight of stairs ~3000 J10 s1 min1 hour

Power is work done per unit time

Average Power:

Instantaneous Power:

1 hp = 746 W1 J

1 s = 1 WattUnits

Work

time

t

WPavg

Fvdt

dxF

dt

dWP

347/8/04

Power in 3-D

For a constant force:

dzFdyFdxFdW zyx

vF

vFvFvFdt

dzF

dt

dyF

dt

dxF

dt

dWP

zzyyxx

zyx

So power will be:

357/8/04

Express Elevator

Say a 900 kg elevator travels 135 floors (400 m) in 40 s. It accelerates at 2.4 m/s2 until it reaches a velocity of 12 m/s.

Find the average power (going up)

W = mgh = (900 kg)(9.8 m/s2)(400 m)

= 3.5 x 106 J

Ws

J

t

WPavg

46

108.840

105.3

m = 900 kg

a = 2.4 m/s2

vmax = 12 m/s

367/8/04

Maximum Instantaneous Power:

P = Fv Maximum just when elevator reaches cruising speed going up

P = (900 kg)(9.8 m/s2+2.4 m/s2)(12 m/s)

= 130,000 W

Express Elevator(continued)

m = 900 kg

a = 2.4 m/s2

vmax = 12 m/sF – mg = ma

a

F

F = m(g+a)

377/8/04

An Application

A car engine can supply some maximum amount of power.

How will the car’s velocity change when it accelerates at constant power?

387/8/04

Acceleration at Constant Power

time

WorkPower timePowerWork

Start from v = 0 and use work-energy theorem:

Acceleration is not constant!

m

Ptv

m

Ptv

Ptmv

2

22

1

2

2

397/8/04

Why Cars Need Gears

There is some maximum power at which the engine can operate. Shifting to a higher gear reduces the force applied to the wheels, allowing for a higher top speed.

F must be larger than the air resistance for the car to continue accelerating.

FvP

407/8/04

Accelerating from 0 to 60

How much average horsepower is needed to accelerate a 1100 kg car from 0 to 60 mph in 3.1 seconds?

In reality, there are transmission losses, etc., so you need much more horsepower to achieve this level of performance.

WKK if

horsepower 711

1027.1

1.3

)/8.26(1100

1.3

)60(1100

5

221

221

W

s

smkg

s

mphkgP

t

mvP

Ptmv2

21

221

417/8/04

Example: (Problem 7.37)

The force (but not the power) to tow a boat at constant velocity is proportional to the speed. If a speed of 4.0 km/h requires 7.5 kW, how much power does a speed of 12 km/hr require?