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Lecture PowerPoints

Chapter 8

Physics for Scientists and Engineers, with Modern

Physics, 4th edition

Giancoli


Chapter 8

Conservation of Energy

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Units of Chapter 8

• Conservative and Nonconservative Forces

• Potential Energy

• Mechanical Energy and Its Conservation

• Problem Solving Using Conservation of Mechanical Energy

• The Law of Conservation of Energy

• Energy Conservation with Dissipative Forces: Solving Problems


Units of Chapter 8

• Gravitational Potential Energy and Escape Velocity

• Power

• Potential Energy Diagrams; Stable and Unstable Equilibrium

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8-1 Conservative and Nonconservative Forces

A force is conservative if:the work done by the force on an object moving from one point to another depends only on the initial and final positions of the object, and is independent of the particular path taken.Example: gravity.

(a) (b)



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m

h

G

:gravityby doneWork

distance fallsObject

Motion Vertical - (a) Case

gF

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12

2

1

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1221

2

1

mghW

yymgdymgW

ddy

dyd

dmgdW

yyhyy

G

y

yG

ll

ll



Another definition of a conservative force:

a force is conservative if the net work done by the force on an object moving around any closed path is zero.

(a) A tiny object moves between points 1 and 2 via two different paths, A and B.

(b) The object makes a round trip, via path A from point 1 to point 2 and via path B back to point 1.

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(a) (b)

Case (a)If force is conservative the same work is involved in moving from 1 to 2 via either path WA = WB

Case (b) WA = -WB Net work = WA - WB = 0



If friction is present, the work done depends not only on the starting and ending points, but also on the path taken. Friction is called a nonconservative force.

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Nonconservative

force

A crate is pushed at constant speed across a rough floor from position 1 to position 2 via two paths, one straight and one curved. The pushing force is always in the direction of motion. (The friction force opposes the motion.) Hence for a constant magnitude pushing force, the work it does is W = FP d, so if d is greater (as for the curved path), then W is greater. The work done does not depend only on points 1 and 2; it also depends on the path taken for a nonconservative force.

PF

frF



Potential energy can only be defined for conservative forces.

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8-2 Potential Energy

An object can have potential energy by virtue of its surroundings.

Familiar examples of potential energy:

• A wound-up spring

• A stretched elastic band

• An object at some height above the ground



In raising a mass m to a height h, the work done by the external force is

We therefore define the gravitational potential energy at a height y above some reference point:

.

.

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This potential energy can become kinetic energy if the object is dropped.

Potential energy is a property of a system as a whole, not just of the object (because it depends on external forces).

If Ugrav = mgy, where do we measure y from?

It turns out not to matter, as long as we are consistent about where we choose y = 0. Only changes in potential energy can be measured.



Example 8-1: Potential energy changes for a roller coaster.

A 1000 kg roller-coaster car moves from point 1 to point 2 and then to point 3.(a) What is the gravitational potential energy at points 2 and 3 relative to point 1? That is, take y = 0 at point 1.(b) What is the change in potential energy when the car goes from point 2 to point 3?(c) Repeat parts (a) and (b), but take the reference point

(y = 0) to be at point 3.

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8-2 Potential EnergyExample 8-1: Potential energy changes for a roller coaster.

J 105.115)(9.81000

m 15 :3Point

J 109.8109.81000

m 10 :2Point

53

13

42

12

mgyU

yyy

mgyU

yyy

A 1000 kg roller-coaster car moves from point 1 to point 2 and then to point 3.

(a) What is the gravitational potential energy at points 2 and 3 relative to point 1? That is, take y = 0 at point 1.


8-2 Potential EnergyExample 8-1: Potential energy changes for a roller coaster.

A 1000 kg roller-coaster car moves from point 1 to point 2 and then to point 3.

(b) What is the change in potential energy when the car goes from point 2 to point 3?

J 102.5109.8105.1Δ

J 105.1 :3Point

J 109.8 :2Point

54523

53

42

UUU

U

U

Potential Energy decreases

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8-2 Potential EnergyExample 8-1: Potential energy changes for a roller coaster.A 1000 kg roller-coaster car moves from point 1 to point 2 and then to point 3.

(c) Repeat parts (a) and (b), but take the reference point ( y = 0) to be at point 3.

J 102.5J 102.50Δ

0 :3Point

J 105.2 :2Point (b)

]definitionby is [this 0

0 :3Point

J 105.2529.81000

m 52 :2Point (a)

5523

3

52

3

33

52

32

UUU

U

U

U

yyy

mgyU

yyy



Example 8-1: Potential energy changes for a roller coaster.

The results from part (c) show that the potential energy depends on the (arbitrary) reference point

The change in potential energy is the same regardless of the reference point

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General definition of gravitational potential energy:

For any conservative force:



A spring has potential energy, called elastic potential energy, when it is compressed. The force required to compress or stretch a spring is:

where k is called the spring constant, and needs to be measured for each spring.

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Then the potential energy is:



In one dimension,

We can invert this equation to find U(x)if we know F(x):

In three dimensions:

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8-3 Mechanical Energy and Its Conservation

If there are no nonconservative forces, the sum of the changes in the kinetic energy and in the potential energy is zero—the kinetic and potential energy changes are equal but opposite in sign.

This allows us to define the total mechanical energy:

And its conservation:

.


8-3 Mechanical Energy and Its Conservation

The principle of conservation of mechanical energy:

If only conservative forces are doing work, the total mechanical energy of a system neither increases nor decreases in any process. It stays constant—it is conserved.

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8-4 Problem Solving Using Conservation of Mechanical Energy

In the image on the left, the total mechanical energy at any point is:

The rock’s potential energy changes to kinetic energy as it falls



Example 8-3: Falling rock.

If the original height of the rock is y1 = h = 3.0 m, calculate the rock’s speed when it has fallen to 1.0 m above the ground.

m/s 6.3)0.10.3(8.922

0

212

222

121

222

12

212

11

2211

yygv

vgygy

mvmgymvmgy

KUKU

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Example 8-4: Roller-coaster car speed using energy conservation.

Assuming the height of the hill is 40 m, and the roller-coaster car starts from rest at the top, calculate (a) the speed of the roller-coaster car at the bottom of the hill, and (b) at what height it will have half this speed. Take y = 0 at the bottom of the hill.




m/s 82408.922

0

212

222

121

222

12

212

11

2211

yygv

vgygy

mvmgymvmgy

KUKU(a)

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8

8

0

If

22

1'2

22'

21

222

121'2'

221'

1

'2222

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g

g

g

vyy

vgygy

vgyvgygy

yyvv(b)



Conceptual Example 8-5: Speeds on two water slides.

Two water slides at a pool are shaped differently, but start at the same height h. Two riders, Paul and Kathleen, start from rest at the same time on different slides. (a) Which rider, Paul or Kathleen, is traveling faster at the bottom? (b) Which rider makes it to the bottom first? Ignore friction and assume both slides have the same path length.

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Conceptual Example 8-5: Speeds on two water slides.

a. The speeds are the same, due to conservation of energy.

b. At every point, Kathleen has less potential energy than Paul, and therefore more kinetic energy. This means her speed is faster (until the end) and she reaches the bottom first.



Which to use for solving problems?

Newton’s laws: best when forces are constant

Work and energy: good when forces are constant; also may succeed when forces are not constant

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Example 8-6: Pole vault.

Estimate the kinetic energy and the speed required for a 70 kg pole vaulter to just pass over a bar 5.0 m high. Assume the vaulter’s center of mass is initially 0.90 m off the ground and reaches its maximum height at the level of the bar itself.



Example 8-6: Pole vault. m = 70 kg

h = 5.0 – 0.9 = 4.1 m

m/s 9.01.48.922

J 28001.48.970

221

ghv

mghmvK

mghUK

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For an elastic force, conservation of energy tells us:

Example 8-7: Toy dart gun.

A dart of mass 0.100 kg is pressed against the spring of a toy dart gun. The spring (with spring stiffness constant k = 250 N/m and ignorable mass) is compressed 6.0 cm and released. If the dart detaches from the spring when the spring reaches its natural length (x = 0), what speed does the dart acquire?



For an elastic force, conservation of energy tells us:

v1 = 0x1 = -0.06 mx2 = 0k = 250 N/mm = 0.1 kg

m/s 3.0

1.0

06.0250

00

221

2

222

1212

1

m

kxv

mvkx

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Example 8-8: Two kinds of potential energy.

A ball of mass m = 2.60 kg, starting from rest, falls a vertical distance h = 55.0 cm before striking a vertical coiled spring, which it compresses an amount Y = 15.0 cm. Determine the spring stiffness constant of the spring. Assume the spring has negligible mass, and ignore air resistance. Measure all distances from the point where the ball first touches the uncompressed spring (y = 0 at this point).



Example 8-8: Two kinds of potential energy.

m = 2.60 kg, starting from rest, h = 55.0 cmY = 15.0 cm.(y = 0 at y2 = 0)

N/m 159015.0

)15.0(55.08.960.222

00

223

31

232

131

3311

y

yymgk

kymgymgy

KUKU

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Example 8-9: A swinging pendulum.This simple pendulum consists of a small bob of mass m suspended by a massless cord of length l. The bob is released (without a push) at t = 0, where the cord makes an angle θ = θ0 to the vertical.

(a) Describe the motion of the bob in terms of kinetic energy and potential energy. Then determine the speed of the bob (b) as a function of position θ as it swings back and forth, and (c) at the lowest point of the swing. (d) Find the tension in the cord,

T. Ignore friction and air resistance.F


Example 8-9: A swinging pendulum.

(a) Describe the motion of the bob in terms of kinetic energy and potential energy.

The speed is zero at extreme points where it has maximum potential energyUmax = mgy0 = mg(l – l cos)

= mgl(1 - cos)

Kinetic energy is maximum at central point

)cos1()cos(max2max2

1max lll mgmgUmvK

At all times: E = U + K

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(b) Speed of the bob as a function of position θ.

0

000

0

02

21

0

000

221

coscos2

coscos)cos(cos

angle of In terms

)(2

E :pointother any At

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yygv

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mgyE

yyt

mgymvUKE

l

lllll

lll



(c) Speed of the bob at the lowest point of the swing.

0cos12

1cos0 :pointlowest At the

θgv

l

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(d) Find the tension in the cord,

T. Ignore friction and air resistance.F

0

0

2

T

2

R

cos2cos3

coscoscos2

cos

cos

mg

gθg

m

gv

mF

mvmamgFF T

ll

l

l


8-5 The Law of Conservation of Energy

Nonconservative, or dissipative, forces:FrictionHeatElectrical energyChemical energyand more

do not conserve mechanical energy. However, when these forces are taken into account, the total energy is still conserved:

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8-5 The Law of Conservation of Energy

The law of conservation of energy is one of the most important principles in physics.

The total energy is neither increased nor decreased in any process. Energy can be transformed from one form to another, and transferred from one object to another, but the total amount remains constant.


8-6 Energy Conservation with Dissipative Forces: Solving Problems

Problem Solving:

1. Draw a picture.

2. Determine the system for which energy will be conserved.

3. Figure out what you are looking for, and decide on the initial and final positions.

4. Choose a logical reference frame.

5. Apply conservation of energy.

6. Solve.

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Example 8-10: Friction on the roller-coaster car.

The roller-coaster car shown reaches a vertical height of only 25 m on the second hill before coming to a momentary stop. It traveled a total distance of 400 m.

Determine the thermal energy produced and estimate the average friction force (assume it is roughly constant) on the car, whose mass is 1000 kg.


Example 8-10: Friction on the roller-coaster car.

m = 1000 kgd = 400 m

J 101.47)2540(8.91000

LossEnergy ProducedEnergy Thermal

5

2121

yymgmgymgy

N 370400

1047.1

LossEnergy 5

frfr

frfr

d

WF

dFW

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Example 8-11: Friction with a spring.

A block of mass m sliding along a rough horizontal surface is traveling at a speed v0 when it strikes a massless spring head-on and compresses the spring a maximum distance X. If the spring has stiffness constant k, determine the coefficient of kinetic friction between block and surface.


Example 8-11: Friction with a spring.

mg

kX

gX

v

mgX

kXmv

mgXkXmv

mgXXFW

mgFF

kXUK

mvK

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Energy Final Energy Initial

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:Force Frictional

&0 :ncompressio maximumAt

hasblock collision ofmoment At

20

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K

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202

1

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8-7 Gravitational Potential Energy and Escape Velocity

Far from the surface of the Earth, the force of gravity is not constant:

The work done on an object moving in the Earth’s gravitational field is given by:



Solving the integral gives:

Because the value of the integral depends only on the end points, the gravitational force is conservative and we can define gravitational potential energy:

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Example 8-12: Package dropped from high-speed rocket.

A box of empty film canisters is allowed to fall from a rocket traveling outward from Earth at a speed of 1800 m/s when 1600 km above the Earth’s surface. The package eventually falls to the Earth. Estimate its speed just before impact. Ignore air resistance.


Example 8-12: Package dropped from high-speed rocket.

2211

1

6

24E

6E

/kgN.m 106.67

m/s 1800

m 101.60

kg 105.98

m 106.38

G

v

h

M

r

m/s 5320

112

Energy Final Energy Initial

EEE

212

E

E

E

E212

1222

1

E

E222

1

E

E212

1

hrrGMvv

hr

GmM

r

GmMmvmv

r

GmMmv

hr

GmMmv

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If an object’s initial kinetic energy is equal to the potential energy at the Earth’s surface, its total energy will be zero. The velocity at which this is true is called the escape velocity; for Earth:

2211

24E

6E

/kgN.m 106.67

kg 105.98

m 106.38

G

M

r

km/s 11.2m/s 101.12

2

0

4

E

Eesc

E

E221

r

GMv

r

GmMmvE


Example 8-13: Escaping the Earth or the Moon.

(a) Compare the escape velocities of a rocket from the Earth and from the Moon.

km/s 37.2m/s 1037.22

:Moon

km/s 11.2m/s 101.122

:Earth

3

M

Mesc

4

E

Eesc

r

GMv

r

GMv

(b) Compare the energies required to launch the rockets. For the Moon, MM = 7.35 x 1022 kg, rM = 1.74 x 106 m, and for Earth, ME = 5.98 x 1024 kg, rE = 6.38 x 106 m.

2237.2

2.1122

esc(Moon)

esc(Earth)

2ese(Moon)2

1

2ese(Earth)2

1

Moon

Earth

v

v

mv

mv

K

K

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8-8 Power

Power is the rate at which work is done.

Average power:

In the SI system, the units of power is the watt:

Instantaneous power:


8-8 Power

Power can also be described as the rate at which energy is transformed:

In the British system, the basic unit for power is the foot-pound per second, but more often horsepower is used:

1 hp = 550 ft·lb/s = 746 W.

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8-8 PowerExample 8-14: Stair-climbing power.

A 60 kg jogger runs up a long flight of stairs in 4.0 s. The vertical height of the stairs is 4.5 m.

(a) Estimate the jogger’s power output in watt and horsepower.

(b) How much energy did this require?

hp 0.88746

660

W6600.4

2600

J26005.48.960

t

WP

mghUW

J 2600Energy (a) From


8-8 Power

Power is also needed for acceleration and for moving against the force of friction.

The power can be written in terms of the net force and the velocity:

If and are in the same direction P = FvF

v

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8-8 PowerExample 8-15: Power needs of a car.

Calculate the power required of a 1400 kg car under the following circumstances: (a) the car climbs a 10° hill (a fairly steep hill) at a steady 80 km/h; and (b) the car accelerates along a level road from 90 to 110 km/h in 6.0 s to pass another car. Assume that the average retarding force on the car is FR = 700 N throughout.


8-8 PowerExample 8-15: Power needs of a car.

Calculate the power required of a 1400 kg car under the following circumstances:

(a) the car climbs a 10° hill (a fairly steep hill) at a steady 80 km/h. Assume that the average retarding force on the car is FR = 700 N throughout.(80 km/h = 80x1000/3600 = 22 m/s)

hp) (91 W 106.8223100

N 310070010sin8.91400

sin

4

Rx

FvP

FmgF

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Example 8-15: Power needs of a car.

Calculate the power required of a 1400 kg car under the following circumstances:

(b) the car accelerates along a level road from 90 km/h to 110 km/h in 6.0 s to pass another car. Assume that the average retarding force on the car is FR = 700 N.90 km/h = 90x1000/3600 = 25.0 m/s110 km/h = 110x1000/3600 = 30.6 m/s

hp) (82 W 106.1230.62000

N 2000700933.01400

m/s 0.9330.6

0.256.30

km/h110at requiredispower maximum the As

4maxmax

R

R

200

FvP

FmaF

maFFFt

vvaatvv

FvP


8-9 Potential Energy Diagrams; Stable and Unstable Equilibrium

This is a potential energy diagram for a particle moving under the influence of a conservative force. Its behavior will be determined by its total energy.

With energy E1, the object oscillates between x3 and x2, called turning points. An object with energy E2 has four turning points; an object with energy E0 is in stable equilibrium. An object at x4 is in unstable equilibrium.

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Summary of Chapter 8

• Conservative force: work depends only on end points

• Gravitational potential energy: Ugrav = mgy.

• Elastic potential energy: Uel = ½ kx2.

• For any conservative force:

• Inverting,



• Total mechanical energy is the sum of kinetic and potential energies.

• Additional types of energy are involved when nonconservative forces act.

• Total energy (including all forms) is conserved.

• Gravitational potential energy:

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• Power: rate at which work is done, or energy is transformed:

or

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