sm chapter8

8Conservation of Energy

CHAPTER OUTLINE

8.1 The Nonisolated System—Conservation of Energy

8.2 The Isolated System8.3 Situations Involving Kinetic Friction8.4 Changes in Mechanical Energy for

Nonconservative Forces8.5 Power

ANSWERS TO QUESTIONS

*Q8.1 Not everything has energy. A rock stationary on the f loor, chosen as the y = 0 reference level, has no mechan-ical energy. In cosmic terms, think of the burnt-out core of a star far in the future after it has cooled nearly to absolute zero.

*Q8.2 answer (c). Gravitational energy is proportional to the mass of the object in the Earth’s fi eld.

*Q8.3 (i) answer b. Kinetic energy is proportional to mass.

(ii) answer c. The slide is frictionless, so v = (2gh)1� 2 in both cases.

(iii) answer a. g for the smaller child and g sin θ for the larger.

*Q8.4 (a) yes: a block slides on the fl oor where we choose y = 0.

(b) yes: a picture on the classroom wall high above the f loor.

(c) yes: an eraser hurtling across the room.

(d) yes: the block stationary on the fl oor.

*Q8.5 answer (d). The energy is internal energy. Energy is never “used up.” The ball fi nally has no elevation and no compression, so it has no potential energy. There is no stove, so no heat is put in. The amount of sound energy is minuscule.

*Q8.6 answer (a). We assume the light band of the slingshot puts equal amounts of kinetic energy into the missiles. With three times more speed, the bean has nine times more squared speed, so it must have one-ninth the mass.

Q8.7 They will not agree on the original gravitational energy if they make different y = 0 choices. They see the same change in elevation, so they do agree on the change in gravitational energy and on the kinetic energy.

Q8.8 Lift a book from a low shelf to place it on a high shelf. The net change in its kinetic energy is zero, but the book-Earth system increases in gravitational potential energy. Stretch a rubber band to encompass the ends of a ruler. It increases in elastic energy. Rub your hands together or let a pearl drift down at constant speed in a bottle of shampoo. Each system (two hands; pearl and shampoo) increases in internal energy.

Q8.9 All the energy is supplied by foodstuffs that gained their energy from the sun.

175

13794_08_ch08_p175-208.indd 17513794_08_ch08_p175-208.indd 175 12/2/06 1:24:39 PM12/2/06 1:24:39 PM

Q8.10 The total energy of the ball-Earth system is conserved. Since the system initially has gravitational energy mgh and no kinetic energy, the ball will again have zero kinetic energy when it returns to its original position. Air resistance will cause the ball to come back to a point slightly below its initial position. On the other hand, if anyone gives a forward push to the ball anywhere along its path, the demonstrator will have to duck.

Q8.11 Let the gravitational energy be zero at the lowest point in the motion. If you start the vibration by pushing down on the block (2), its kinetic energy becomes extra elastic potential energy in the spring (Us). After the block starts moving up at its lower turning point (3), this energy becomes both kinetic energy (K ) and gravitational potential energy (U g), and then just gravitational energy when the block is at its greatest height (1). The energy then turns back into kinetic and elastic potential energy, and the cycle repeats.

*Q8.12 We have (1� 2)mv2 = µkmgd so d = v2�2µ

kg. The quantity v2�µ

k controls the skidding distance.

In the cases quoted respectively, this quantity has the numerical value (a) 5 (b) 2.5 (c) 1.25 (d) 20 (e) 10 (f ) 5. In order the ranking is then d > e > f = a > b > c.

*Q8.13 Yes, if it is exerted by an object that is moving in our frame of reference. The fl at bed of a truck exerts a static friction force to start a pumpkin moving forward as it slowly starts up.

*Q8.14 (a) A campfi re converts chemical energy into internal energy, within the system wood-plus-oxygen, and before energy is transferred by heat and electromagnetic radiation into the surroundings. If all the fuel burns, the process can be 100% effi cient. Chemical-energy-into-internal-energy is also the conversion as iron rusts, and it is the main conversion in mamma-lian metabolism.

(b) An escalator motor converts electrically transmitted energy into gravitational energy. As the system we may choose motor-plus-escalator-and-riders. The effi ciency could be say 90%, but in many escalators plenty of internal energy is another output. A natural process, such as atmospheric electric current in the Earth’s aurora borealis raising the temperature of a particu-lar parcel of air so that the surrounding air buoys it up, could produce the same electrically-transmitted-to-gravitational energy conversion with low effi ciency.

(c) A diver jumps up from a diving board, setting it vibrating temporarily. The material in the board rises in temperature slightly as the visible vibration dies down, and then the board cools off to the constant temperature of the environment. This process for the board-plus-air system can have 100% effi ciency in converting the energy of vibration into energy trans-ferred by heat. The energy of vibration is all elastic energy at instants when the board is momentarily at rest at turning points in its motion. For a natural process, you could think of the branch of a palm tree vibrating for a while after a coconut falls from it.

(d) Some of the sound energy in a shout becomes a tiny bit of work done on a listener’s ear; most of the mechanical-wave energy becomes internal energy as the sound is absorbed by all the surfaces it falls upon. We would also assign low effi ciency to a train of water waves doing work to make a linear pile of shells at the high-water mark on a beach.

(e) A demonstration solar car takes in electromagnetic-wave energy in sunlight and turns some fraction of it temporarily into the car’s kinetic energy. A much larger fraction becomes internal energy in the solar cells, battery, motor, and air pushed aside. Perhaps with somewhat higher net effi ciency, the pressure of light from a newborn star pushes away gas and dust in the nebula surrounding it.

176 Chapter 8

FIG. Q8.11


Conservation of Energy 177

*Q8.15 (a) original elastic potential energy into fi nal kinetic energy

(b) original chemical energy into fi nal internal energy

(c) original internal energy in the batteries into fi nal internal energy, plus a tiny bit of outgoing energy transmitted by mechanical waves

(d) original kinetic energy into fi nal internal energy in the brakes

(e) heat input from the lower layers of the Sun, into energy transmitted by electromagnetic radiation

(f ) original chemical energy into fi nal gravitational energy

*Q8.16 Answer (k). The static friction force that each glider exerts on the other acts over no distance. The air track isolates the gliders from outside forces doing work. The gliders-Earth system keeps constant mechanical energy.

Q8.17 The larger engine is unnecessary. Consider a 30-minute commute. If you travel the same speed in each car, it will take the same amount of time, expending the same amount of energy. The extra power available from the larger engine isn’t used.

SOLUTIONS TO PROBLEMS

Section 8.1 The Nonisolated System—Conservation of Energy

*P8.1 (a) The toaster coils take in energy by electrical transmission. They increase in internal energy and put out energy by heat into the air and energy by electromagnetic radiation as they start

to glow. int ET ER∆E Q T T= + +

(b) The car takes in energy by mass transfer. Its fund of chemical potential energy increases.As it moves, its kinetic energy increases and it puts out work on the air, energy by heat in the exhaust, and a tiny bit of energy by mechanical waves in sound.

int MW MT∆ ∆ ∆K U E W Q T T+ + = + + +

(c) You take in energy by mass transfer. Your fund of chemical potential energy increases. You are always putting out energy by heat into the surrounding air. MT∆U = +Q T

(d) Your house is in steady state, keeping constant energy as it takes in energy by electricaltransmission to run the clocks and, we assume, an air conditioner. It absorbs sunlight, taking in energy by electromagnetic radiation. The exterior plenum of the air conditioner takes in cooler air and puts it out as warmer air, transferring out energy by mass transfer.

MT ET ER0 = + + +Q T T T

Section 8.2 The Isolated System

P8.2 (a) One child in one jump converts chemical energy into mechanical energy in the amount that her body has as gravitational energy at the top of her jump:

mgy = ( )( ) =36 9 81 0 25 88 3kg m s m J2. . .

For all of the jumps of the children the energy is 12 1 05 10 88 3 1 11 106 9. . .×( ) = ×J J .

(b) The seismic energy is modeled as E = × = ×0 01

1001 11 10 1 11 109 5.. .J J, making the Richter

magnitude log .

.

log . .

.

. .

.

E − = × − = − =4 8

1 5

1 11 10 4 8

1 5

5 05 4 8

1 5

5

00 2. .


178 Chapter 8

P8.3 U K U Ki i f f+ = + : mgh mg R m+ = ( ) +0 21

22v

g R g R3 50 21

22.( ) = ( ) + v

v = 3 00. gR

F mR∑ =v2

: n mg mR

+ =v2

n mR

g mgR

Rg mg

n

= −⎡⎣⎢

⎤⎦⎥

= −⎡⎣⎢

⎤⎦⎥

=

=

v2 3 002 00

2 0

..

. 00 5 00 10 9 80

0 098 0

3. .

.

×( )( )=

− kg m s

N downwar

2

dd

P8.4 (a) ∆ ∆K W W mg h mgA B g( ) = = = = −( )→ ∑ 5 00 3 20. .

1

2

1

29 80 1 80

5 94

2 2m m mB A

B

v v

v

− = ( )( )

=

. .

. m s

Similarly, v vC A g= + −( ) =2 2 5 00 2 00 7 67. . . m s

(b) W mgg A C→= ( ) =3 00 147. m J

P8.5 From conservation of energy for the block-spring-Earth system,

U Ugt si=

or

0 250 9 801

25 000 0 100. . .kg m s N m2( )( ) = ⎛

⎝⎞⎠ ( )h mm( )2

This gives a maximum height h = 10 2. m .

P8.6 (a) The force needed to hang on is equal to the force F the trapeze bar exerts on the performer.

From the free-body diagram for the performer’s body,as shown,

F mg m− =cosθ v2

� or

F mg m= +cosθ v2

�

Apply conservation of mechanical energy of the performer-Earth system as the performer moves between the starting point and any later point:

mg mg mi� � � �−( ) = −( ) +cos cosθ θ 1

22v

Solve for mv2

� and substitute into the force equation to obtain F mg i= −( )3 2cos cosθ θ .

FIG. P8.3

5.00 m

A

B

C

2.00 m3.20 m

FIG. P8.4

FIG. P8.5

FIG. P8.6

continued on next page



(b) At the bottom of the swing, θ = 0°

so

F mg

F mg mg

i

i

= −( )= = −( )

3 2

2 3 2

cos

cos

θ

θ which gives

θi = 60 0. °

P8.7 Using conservation of energy for the system of the Earth and the two objects

(a) 5 00 4 00 3 00 4 001

25 0. . . . .kg m kg m( ) ( ) = ( ) ( ) +g g 00 3 00 2+( ). v

v = =19 6 4 43. . m s

(b) Now we apply conservation of energy for the system of the 3.00 kg object and the Earth during the time interval between the instant when the string goes slack and the instant at which the 3.00 kg object reaches its highest position in its free fall.

1

23 00 3 00

1 00

4 00

2. .

.

.max

( ) = =

=

=

v mg y g y

y

y

∆ ∆

∆ m

m ++ =∆y 5 00. m

P8.8 We assume m m1 2>

(a) m gh m m m gh1 1 22

2

1

2= +( ) +v

v =−( )+( )

2 1 2

1 2

m m gh

m m

(b) Since m2 has kinetic energy 1

2 22m v , it will rise an additional height ∆h determined from

m g h m2 221

2∆ = v

or from (a),

∆hg

m m h

m m= =

−( )+( )

v21 2

1 22

The total height m2 reaches is h hm h

m m+ =

+∆

2 1

1 2

.

FIG. P8.7


180 Chapter 8

P8.9 The force of tension and subsequent force of compression in the rod do no work on the ball, since they are perpendicular to each step of displacement. Consider energy conservation of the ball-Earth system between the instant just after you strike the ball and the instant when it reaches the top. The speed at the top is zero if you hit it just hard enough to get it there.

K U K Ui gi f gf+ = + : 1

20 0 22m mg Liv + = + ( )

v

v

i

i

gL= = ( )( )

=

4 4 9 80 0 770

5 49

. .

. m s

P8.10 (a) K U K Ui gi f gf+ = +

1

20

1

21

2

1

2

1

2

2 2

2 2 2

m m mgy

m m m

i f f

xi yi xf

v v

v v v

+ = +

+ = ++ mgyf

Note that we have used the Pythagorean theorem to express the original kinetic energy in terms of the velocity components. Kinetic energy itself does not have components.

Now v vxi xf= , so for the fi rst ball

ygfyi= =

( )( ) = ×

v2 24

2

1 000 37 0

2 9 801 85 10

sin .

..

°m

and for the second

yf =( )

( ) = ×1 000

2 9 805 10 10

24

.. m

(b) The total energy of each is constant with value

1

220 0 1 000 1 00 10

2 7. .kg m s J( )( ) = ×

P8.11 (a) For a 5-m cord the spring constant is described by F kx= ,mg k= ( )1 5. m . For a longer cord of length L the stretch distance is longer so the spring constant is smaller in inverse proportion:

kL

mgmg L

K U U K U Ug s i g s f

= =

+( ) = + +( )

53 33

0

m

1.5 m.

+

++ + = + +

−( ) = =

mgy mgy kx

mg y y kx

i f f

i f f

0 01

21

2

1

23

2

2 .333 2mg

Lx f

here y y L xi f f− = = +55 m

55 01

23 33 55 0

55 0 5 04 10

2

3

. . .

. .

m m

m m

L L

L

= −( )= × 22 m− +

= − + × =

=

183 1 67

0 1 67 238 5 04 10 0

2

2

2 3

L L

L L

L

.

. .

338 238 4 1 67 5 04 10

2 1 67

238 152

3

2 3± − ( ) ×( )( ) = ±. .

. .33325 8= . m

only the value of L less than 55 m is physical.

L

vi

initialL

final

FIG. P8.9

initial final

FIG. P8.11(a)




(b) kmg

= 3 3325 8

.. m

x x fmax . . .= = =55 0 25 8 29 2m m m−

F ma∑ = + − =kx mg mamax

3 3325 8

29 2

2 77 27 1

..

.

. .

mgmg ma

a g

mm

m s2

− =

= =

P8.12 When block B moves up by 1 cm, block A moves down by 2 cm and the separation becomes

3 cm. We then choose the fi nal point to be when B has moved up by h

3 and has speed

vA

2.

Then A has moved down 2

3

h and has speed vA:

K K U K K U

m m

g i g fA B A B

A2 A

+ +( ) = + +( )+ + = + ⎛0 0 0

1

2

1

2 2v

v⎝⎝

⎞⎠ + −

=

=

2

3

2

3

3

5

8

8

15

mgh mg h

mghm

gh

v

v

A2

A

Section 8.3 Situations Involving Kinetic Friction

P8.13 F may y∑ = : n − =392 0N

n

f nk k

== = ( )( ) =392

0 300 392 118

N

N Nµ .

(a) W F rF = = ( )( ) =∆ cos . cosθ 130 5 00 0 650° J

(b) ∆ ∆E f xkint J= = ( )( ) =118 5 00 588.

(c) W n rn = = ( )( ) =∆ cos . cosθ 392 5 00 90 0°

(d) W mg rg = = ( )( ) −( ) =∆ cos . cosθ 392 5 00 90 0°

(e) ∆ ∆K K K W Ef i= − = −∑ other int

1

20 650 588 0 0 62 02m fv − = − + + =J J J.

(f ) v ffK

m= = ( )

=2 2 62 0

40 01 76

.

..

J

kgm s

FIG. P8.13


182 Chapter 8

P8.14 (a) W kx kxs i f= −

= ( ) ×( ) − =−

1

2

1

21

2500 5 00 10 0 0 62

2 2

2 2. . 55 J

W m m ms f i f= − = −1

2

1

2

1

202 2 2v v v

so

v f

W

m=

( )

= ( )=

∑2

2 0 625

2 000 791

.

..m s m s

(b) 1

2

1

22 2m f x W mi k s fv v− + =∆

0 0 350 2 00 9 80 0 050 0 0 6251

2− ( )( )( )( ) + =. . . . .J J mvv

v

v

f

f

f

2

20 2821

22 00

2 0 282

2 00

. .

.

.

J kg

m

= ( )

= ( )ss m s= 0 531.

P8.15 (a) W mgg = +( )�cos .90 0° θ

Wg = ( )( )( )

= −

10 0 9 80 5 00 110

168

2. . . coskg m s m °

JJ

(b) f n mgk k k= =µ µ θcos

∆∆

E f mg

E

k kint

int m

= == ( )( )

� �µ θcos

. . .5 00 0 400 10 00 9 80 20 0

184

( )( )=

. cos . °

J

(c) W FF = = ( )( ) =� 100 5 00 500. J

(d) ∆ ∆ ∆K W E W W EF g= − = + − =∑ other int int J148

(e) ∆K m mf i= −1

2

1

22 2v v

v vf i

K

m= ( )

+ = ( )+ ( ) =

2 2 148

10 01 50 5 652 2∆

.. . m s

*P8.16 (i) In (a), (kd 2)1� 2 and (mgd )1� 2 both have the wrong units for speed. In (b) (µ kg)1� 2 has the

wrong units. In (c), (kd�m)1� 2 has the wrong units. In (f ) both terms have the wrong units. The answer list is a, b, c, f .

(ii) As k increases, friction becomes unimportant, so we should have (1� 2)kd 2 = (1� 2)mv2 andv = (kd 2�m)1� 2. Possibilities g, i, and j do not have this limit.

(iii) As µ k goes to zero, as in (ii), we should have v = (kd 2�m)1� 2. Answer d does not have this limit.

FIG. P8.15


FIG. P8.14



(iv) (e) cannot be true because the friction force is proportional to µ

k and not µ

k2. And (k) can-

not be true because the presence of friction will reduce the speed compared to the µ k = 0

case, and not increase the speed.

(v) If the spring force is strong enough to produce motion against static friction and if the spring energy is large enough to make the block slide the full distance d, the continuity equation for energy gives

(1� 2) kd 2 + µ kmgd cos 180° = (1� 2) mv2

This turns into the correct expression h .

(vi) We have (kd 2�m − 2 µ k gd )1/ 2 = [18 (0.12)2�0.25 − 2 (0.6)(9.8)(0.12)]1� 2 = [1.04 − 1.41]1� 2

The expression gives an imaginary answer because the spring does not contain enough energy in this case to make the block slide the full distance d.

P8.17 vi = 2 00. m s µk = 0 100.

K f x W Ki k f− + =∆ other : 1

202m f xi kv − =∆

1

22m mg xi kv = µ ∆ ∆x

gi

k

= =( )( )( ) =

v2 2

2

2 00

2 0 100 9 802 04

µ.

. ..

m sm

Section 8.4 Changes in Mechanical Energy for Nonconservative Forces

P8.18 (a) U K K Uf i f i= − + U f = − + =30 0 18 0 10 0 22 0. . . . J

E = 40 0. J

(b) Yes, ∆ ∆ ∆E K Umech = + is not equal to zero, some nonconservative force or forces must act. For conservative forces ∆ ∆K U+ = 0.

P8.19 U K E U Ki i f f+ + = +∆ mech: m gh fh m m2 1

22

21

2

1

2− = +v v

f n m g= =µ µ 1

m gh m gh m m2 1 1 2

21

2− = +( )µ v

v2 2 1

1 2

2=

−( )( )+

m m hg

m m

µ

v =( )( ) − ( )2 9 80 1 50 5 00 0 400 3 00. . . . .m s m kg kg2 ⎡⎡⎣ ⎤⎦ =

8 003 74

..

kgm s

FIG. P8.19


184 Chapter 8

P8.20 The distance traveled by the ball from the top of the arc to the bottom is π R. The work done by the non-conservative force, the force exerted by the pitcher, is

∆ ∆E F r F R= = ( )cos0° π

We shall assign the gravitational energy of the ball-Earth system to be zero with the ball at the bottom of the arc.

Then

∆E m m mgy mgyf i f imech = − + −1

2

1

22 2v v

becomes1

2

1

22 2m m mgy F Rf i iv v= + + ( )π

or

v vf i igyF R

m= + + ( )

= ( ) + ( )( ) +2 222

15 0 2 9 80 1 202π

. . .330 0 0 600

0 250

. .

.

( ) ( )π

v f = 26 5. m s

P8.21 (a) ∆K m mf i i= −( ) = − = −1

2

1

21602 2 2v v v J

(b) ∆U mg= ( ) =3 00 30 0 73 5. sin . .m J°

(c) The mechanical energy converted due to friction is 86.5 J

f = =86 5

28 8.

.J

3.00 mN

(d) f n mgk k= = =µ µ cos . .30 0 28 8° N

µk = ( )( ) =28 8

9 80 30 00 679

.

. cos ..

N

5.00 kg m s2 °

P8.22 Consider the whole motion: K U E K Ui i f f+ + = +∆ mech

(a) 01

20

80 0 9 80

1 1 2 22+ − − = +

( )

mgy f x f x mi f∆ ∆ v

. .kg m s22 m N m N m( ) − ( )( ) − ( )( ) =1 000 50 0 800 3 600 200.11

280 0

784 000 40 000 720 0001

2

2. kg

J J J

( )

− − =

v f

880 0

2 24 000

80 024 5

2.

..

kg

J

kgm s

( )

=( )

=

v

v

f

f

(b) Yes. This is too fast for safety.

(c) Now in the same energy equation as in part (a), ∆x2 is unknown, and ∆ ∆x x1 21 000= −m :

784 000 50 0 1 000 3 6001

2 2J N m N− ( ) −( ) − ( ) =. ∆ ∆x x22

80 0 5 00

784 000 50 000 3 550

2. .kg m s

J J

( )( )

− − NN J

J

3 550 Nm

( ) =

= =

∆

∆

x

x

2

2

1 000

733 000206

FIG. P8.21




(d) Really the air drag will depend on the skydiver’s speed. It will be larger than her 784 N weight only after the chute is opened. It will be nearly equal to 784 N before she opens the chute and again before she touches down, whenever she moves near terminal speed.

P8.23 (a) K U E K Ui f+( ) + = +( )∆ mech :

01

2

1

20

1

28 00 5 00 10

2 2

2

+ − = +

( ) ×( )−

kx f x m∆ v

. .N m m22 2 33 20 10 0 150

1

25 30 10− ×( )( ) = ×( )− −. . .N m kg v22

3

3

2 5 20 10

5 30 101 40v =

×( )×

=−

−

.

..

J

kgm s

(b) When the spring force just equals the friction force, the ball will stop speeding up. Here�Fs kx= ; the spring is compressed by

3 20 10

0 4002.

.×

=− N

8.00 N mcm

and the ball has moved

5 00 0 400 4 60. . .cm cm cm from the start.− =

(c) Between start and maximum speed points,w

1

2

1

2

1

21

28 00 5 00 10 3

2 2 2

2 2

kx f x m kxi f− = +

×( ) −−

∆ v

. . .. .

. .

20 10 4 60 10

1

25 30 10

1

28

2 2

3 2

×( ) ×( )

= ×( ) +

− −

− v 000 4 00 10

1 79

3 2.

.

×( )=

−

v m s

P8.24 (a) There is an equilibrium point wherever the graph of potential energy is horizontal:

At r = 1.5 mm and 3.2 mm, the equilibrium is stable.At r = 2.3 mm, the equilibrium is unstable.A particle moving out toward r → ∞ approaches neutral equilibrium.

(b) The system energy E cannot be less than −5.6 J. The particle is bound if − ≤ <5 6 1. J JE .

(c) If the system energy is −3 J, its potential energy must be less than or equal to −3 J. Thus,

the particle’s position is limited to 0 6 3 6. .mm mm≤ ≤r .

(d) K + U = E. Thus, K E Umax min . . .= − = − − −( ) =3 0 5 6 2 6J J J .

(e) Kinetic energy is a maximum when the potential energy is a minimum, at r = 1 5. mm .

(f ) − + =3 1J JW . Hence, the binding energy is W = 4 J .


186 Chapter 8

P8.25 (a) The object moved down distance 1 20. m + x . Choose y = 0 at its lower point.

K U U E K U U

mgy

i gi si f gf sf

i

+ + + = + +

+ + + = + +

∆ mech

0 0 0 0 011

2

1 50 9 80 1 201

2320

2kx

x. . .kg m s m N2( )( ) +( ) = mm

N m N J

N

( )

= ( ) − ( ) −

=

x

x x

x

2

20 160 14 7 17 6

14 7

. .

. ±± −( ) − ( ) − ⋅( )( )

14 7 4 160 17 6

2 160

2. .N N m N m

N m

x ==±14 7 107. N N

320 N m

The negative root tells how high the object will rebound if it is instantly glued to the spring. We want

x = 0 381. m

(b) From the same equation,

1 50 1 63 1 201

2320

0

2. . .kg m s m N m2( )( ) +( ) = ( )x x

== − −160 2 44 2 932x x. .

The positive root is x = 0 143. m .

(c) The equation expressing the energy version of the nonisolated system model has one more term:

mgy f x kx

x

i − =

( )( ) +( )

∆ 1

2

1 50 9 80 1 20

2

. . .kg m s m2 −− +( ) = ( )

+

0 700 1 201

2320

17 6 14 7

2. .

. .

N m N m

J

x x

N J N N mx x x

x x

− − =

− −

0 840 0 700 160

160 14 0 1

2

2

. .

. 66 8 0

14 0 14 0 4 160 16 8

3200 37

2

.

. . .

.

=

=± ( ) − ( ) −( )

=

x

x 11 m

P8.26 The boy converts some chemical energy in his muscles into kinetic energy. During this conversion, the energy can be measured as the work his hands do on the wheels.

K U U f x K

K U W

i gi chemical i k f

i gi hands on

+ + − =

+ + − −

, ∆

wwheels k f

i i byboy k

f x K

m m y W f x m

− =

+ + − =

∆

∆12

2 12v vg ff

2

or

W m mgy f xbyboy f i i= −( ) − +1

22 2v v ∆

Wbyboy = ( ) ( ) − ( )⎡⎣ ⎤⎦ − ( )1

247 0 6 20 1 40 47 0 92 2. . . . .880 2 60 41 0 12 4

168

( )( ) + ( )( )

=

. . .

Wbyboy J

FIG. P8.26



P8.27 (a) Let m be the mass of the whole board. The portion on the rough surface has mass mx

L. The

normal force supporting it is mxg

L and the frictional force is

µkmgx

Lma= . Then

agx

Lk=

µopposite to the motion. .

(b) In an incremental bit of forward motion dx, the kinetic energy converted into internal

energy is f dxmgx

Ldxk

k=µ

. The whole energy converted is

1

2 2 22

0

2

0

mmgx

Ldx

mg

L

x mgL

gL

kL

k

L

k

k

v

v

= = =

=

∫µ µ µ

µ

Section 8.5 Power

P8.28 Pav

kg m s= = = = ( )

×W

t

K

t

m

tf

∆ ∆ ∆v2 2

2

0 875 0 620

2 21

. .

1108 013−( ) =

sW.

P8.29 Power =W

t

P = =( )( )

=mgh

t

700 10 0

8 00875

N m

sW

.

.

*P8.30 (a) The moving sewage possesses kinetic energy in the same amount as it enters and leaves the pump. The work of the pump increases the gravitational energy of the sewage-Earth system. We take the equation K U W K Ui gi f gf+ + = +pump , subtract out the K terms, and choose Ugi = 0 at the bottom of the sump, to obtain W mgyfpump = . Now we differentiate through with respect to time:

Ppump

3kg m L d

= =

= ×

∆∆

∆∆

m

tgy

V

tgy

f fρ

1050 1 89 106.(( )⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

1

1000

1 9 80m

L

d

86 400 s

m

s

3

2

.⎛⎛⎝⎜

⎞⎠⎟

( )= ×

5 49

1 24 103

.

.

m

W

(b) efficien yuseful output work

total input woc =

rrk

useful output work

total input work=

=

∆∆

t

t

mmechanical output power

input electric powerr

kW

5.90 kW= = =1 24

0 209 20 9.

. . %

The remaining power, 5 90 1 24 4 66. . .− =kW kW is the rate at which internal energy is injected into the sewage and the surroundings of the pump.

Dave Barry attended the January dedication of the pumping station and was the featured speaker at a festive potluck supper to which residents of the different Grand Forks sewer districts brought casseroles, Jell-O salads, and “bars” for dessert.


188 Chapter 8

P8.31 A 1 300-kg car speeds up from rest to 55.0 mi � h = 24.6 m �s in 15.0 s. The output work of the engine is equal to its fi nal kinetic energy,

1

21 300 24 6 390

2kg m s kJ( )( ) =.

with power P = 390 000104J

15.0 sW~ around 30 horsepower.

P8.32 (a) The distance moved upward in the fi rst 3.00 s is

∆y t= =+⎡

⎣⎢⎤⎦⎥( ) =v

0 1 75

23 00 2 63

.. .

m ss m

The motor and the earth’s gravity do work on the elevator car:

1

2180

1

21

265

2 2m W mg y m

W

i fv v+ + =

=

motor

motor

∆ cos °

00 1 75 0 650 2 63 1 772

kg m s kg m( )( ) − + ( ) ( ) = ×. . .g 1104 J

Also, W t=P so P = = × = × =W

t

1 77 105 91 10 7 92

43.

. .J

3.00 sW hp .

(b) When moving upward at constant speed v =( )1 75. m s the applied force equals the weight kg m s N2= ( )( ) = ×650 9 80 6 37 103. . . Therefore,

P = = ×( )( )= × =Fv 6 37 10 1 75 1 11 10 14 93 4. . . .N m s W hhp

P8.33 energy power time= × For the 28.0 W bulb:

Energy used = ( ) ×( ) =28 0 1 00 10 2804. .W h kilowattt hrs⋅

total cost kWh kWh= + ( )( ) =$ . $ . $ .17 00 280 0 080 39 440

For the 100 W bulb:

Energy used W h ki= ( ) ×( ) = ×100 1 00 10 1 00 104 3. . llowatt hrs⋅

# bulb usedh

750 h bulb=

×=

1 00 1013 3

4..

total cost kWh= ( ) + ×( )13 3 0 420 1 00 10 0 083. $ . . $ . 00 85 60kWh( ) = $ .

Savings with energy-efficient bulb = −$ . $85 60 339 40 46 2. $ .= .

P8.34 The useful output energy is

120 1 0 60

120 3 600

Wh

W s

−( ) = −( ) =

=

. mg y y F y

y

f i g∆

∆ (( )⋅

⎛⎝⎜

⎞⎠⎟

⋅⎛⎝⎜

⎞⎠⎟ =

0 40

890194

.

N

J

W s

N m

Jm



P8.35 The energy of the car is E m mgy= +1

22v

E m mgd= +1

22v sinθ where d is the distance it has moved along the track.

P = = +dE

dtm

d

dtmgv

vv sinθ

(a) When speed is constant,

P = mgv sin = ( )( ).θ 950 2 20kg 9.80 m s m s2 sin =30 1° ..02 104× W

(b) d

dta

v= =

−=

2 2 0

120 183

..

m s

sm s2

Maximum power is injected just before maximum speed is attained:

P = +m a mgv v sin = ( )( )+. . .θ 950 2 2 0 183 1 0kg m s m s2 22 104× ×W= 1.06 10 W4

(c) At the top end,

1

2950 2 20 9 802 2

m mgdv + = ( ) + ( )sin . .θ kg1

2m s m s2 11 250 30 5 82 106m Jsin .°⎛

⎝⎜⎞⎠⎟ = ×

P8.36 (a) Burning 1 lb of fat releases energy

19 4 186

lb454 g

1 lb

kcal

1 g

J

1 k⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ ccal

J⎛⎝⎜

⎞⎠⎟ = ×1 71 107.

The mechanical energy output is

1 71 10 0 207. . cos×( )( ) =J nF r∆ θ

Then 3 42 10 06. cos× =J nmg y∆ °

3 42 10 50 9 8 80 0 1506. . .× = ( )( )( )J kg m s steps2n m

J J

( )

× = ×( )3 42 10 5 88 106 3. .n

where the number of times she must climb the steps is n =××

=3 42 10

5 88 10582

6

3

.

.

J

J.

This method is impractical compared to limiting food intake.

(b) Her mechanical power output is

P = = × = =⎛W

t

5 88 1090 5 90 5

13.. .

J

65 sW W

hp

746 W⎝⎝⎜⎞⎠⎟

= 0 121. hp


190 Chapter 8

Additional Problems

P8.37 (a) K U K Ug A g B+( ) = +( )

01

202+ = +mgy mA Bv vB Agy= = ( ) =2 2 9 8 6 3 11 1. . .m s m m s2

(b) arc = = ( ) =v2 2

11 1

6 319 6

.

..

m s

mm s up2

(c) F may y∑ = + − =n mg maB c

nB = +( ) = ×76 9 8 19 6 2 23 103kg m s m s N up2 2. . .

(d) We compute the amount of chemical energy converted into mechanical energy as

W F r= = × ( ) = ×∆ cos . . cos .θ 2 23 10 0 450 0 1 01 103 3N m ° J

(e) K U U K Ug chemical B g D+ +( ) = +( )

1

20 1 01 10

1

21

276

2 3 2m m mg y yB D D Bv v+ + × = + −( ). J

kg 111 1 1 01 101

276 76 9 8

2 3 2. . .m s J kg kg( ) + × = +vD mm s m

J J

kg

2( )× − ×( )

=

6 3

5 70 10 4 69 10 2

76

3 3

.

. .vD == 5 14. m s

(f ) K U K Ug D g E+( ) = +( ) where E is the apex of his motion

1

20 02m mg y yD E Dv + = + −( ) y y

gE DD− = = ( )

( ) =v2 2

2

5 14

2 9 81 35

.

..

m s

m sm2

(g) Consider the motion with constant acceleration between takeoff and touchdown. The time is the positive root of

y y t a t

t

f i yi y= + +

− = + + −

v1

2

2 34 0 5 141

29 8

2

. . .m m s mm s2( )− − =

=± −

t

t t

t

2

2

2

4 9 5 14 2 34 0

5 14 5 14 4 4 9

. . .

. . .(( ) −( )=

2 34

9 81 39

.

.. s

*P8.38 (a) Yes, the total mechanical energy is constantt. The originally hanging block loses gravi-tational energy, which is entirely converted into kinetic energy of both blocks.

(b) energy at release = energy just before hitting fl oor m

2gy = (1� 2) (m

1 + m

2)v2

v = [2m2gy �(m

1 + m

2)]1� 2 = [2(1.90 kg)(9.8 m �s2)0.9 m �5.4 kg]1� 2 = 2.49 m/s

(c) No. The kinetic energy of the impacting block turns into internal energy. But mechanical energy is conserved for the 3.50-kg block with the Earth in this block’s projectile motion.




(d) For the 3.5-kg block from when the string goes slack until just before the block hits the fl oor

(1� 2) (m2)v2 + m

2gy = (1� 2) (m

2) v

d 2

vd = [2gy + v2]1� 2 = [2(9.8 m �s2)1.2 m + (2.49 m �s)2]1� 2 = 5.45 m/s

(e) The 3.5-kg block takes this time in fl ight to the fl oor: from y = (1� 2) gt2 we havet = [2(1.2)�9.8]1� 2 = 0.495 s. Its horizontal component of displacement at impact is thenx = v

d t = (2.49 m �s)(0.495 s) = 1.23 m .

(f ) No. With the hanging block fi rmly stuck, the string pulls radially on the 3.5-kg block, doing no work on it.

(g) The force of static friction cannot be larger than µsn = (0.56)(3.5 kg)(9.8 m �s2) = 19.2 N.

The hanging block tends to produce string tension (1.9 kg)(9.8 m �s2) = 18.6 N. Then the force of static friction on the 3.5-kg block is less than its maximum value, being

18.6 N to the left .

(h) A little push is required , because 18.6 N is less than 19.2 N. The motion begins with

negligible speed, so the calculated final speeds are still accuraate .

P8.39 (a) x t t= + 2 00 3.

Therefore,

v

v

= = +

= = ( ) +( ) =

dx

dtt

K m t

1 6 00

1

2

1

24 00 1 6 00

2

2 2 2

.

. . 22 00 24 0 72 02 4. . .+ +( )t t J

(b) ad

dtt= = ( )v

12 0. m s2

F ma t t= = ( ) = ( )4 00 12 0 48 0. . . N

(c) P = = ( ) +( )= +( )F t t t tv 48 0 1 6 00 48 0 2882 3. . . W

(d) W dt t t dt= = +( ) =∫ ∫P0

2 003

0

2 00

48 0 288 1250. .

. J

*P8.40 (a) Simplifi ed, the equation is 0 9 700 450 8 1 3952= ( ) − ( ) − ⋅N m N N mx x. . Then

xb b ac

a=

− ± −=

± ( ) − ( )2 24

2

450 8 450 8 4 9 700. .N N N m −− ⋅( )( )

=±

1 395

2 9 700

450 8 7 370

N m

N m

N N

19 400

.

N mm or m= −0 403 0 357. .

(b) One possible problem statement: From a perch at a height of 2.80 m above the top of the

pile of mattresses, a 46.0-kg child jumps nearly straight upward with speed 2.40 m �s. The mattresses behave as a linear spring with force constant 19.4 kN�m. Find the maximum amount by which they are compressed when the child lands on them. Physical meaning: The positive value of x represents the maximum spring compression. The negative value represents the maximum extension of the equivalent spring if the child sticks to the top of the mattress pile as he rebounds upward without friction.


192 Chapter 8

P8.41 (a) v = = − +( )

=

∫ ∫a dt t t t dt

t

t t

0

2 3

0

2

1 16 0 21 0 24

1 162

. . .

. −− + = − +0 213

0 244

0 58 0 07 0 063 4

0

2 3 4. . . . .t t

t t tt

At t = 0, vi = 0. At t = 2 5. s,

v f = ( )( ) − ( )( ) +0 58 2 5 0 07 2 5 02 3. . . . .m s s m s s3 4 006 2 5 4 88

01

2

4

2

m s s m s5( )( ) =

+ =

+ = =

. .

K W K

W m

i f

fv11

21160 4 88 1 38 10

2 4kg m s J. .( ) = ×

(b) At t = 2 5. s ,

a = ( ) − ( )( ) +1 16 2 5 2 5 0 2402. . . .m s s 0.210 m s s3 4 m s s m s5 2( )( ) =2 5 5 343. .

Through the axles the wheels exert on the chassis force

F ma∑ = = = ×1160 5 34 6 19 103kg m s N2. .

and inject power

P = = × ( )= ×F v 6 19 10 4 88 3 02 103 4. . .N m s W

P8.42 (a) ∆ ∆E K m f iint = − = − −( )1

22 2v v : ∆Eint kg m s= − ( ) ( ) − ( )( )( ) =

1

20 400 6 00 8 00 52 2 2

. . . ..60 J

(b) ∆ ∆E f r mg rkint = = ( )µ π2 : 5 60 0 400 9 80 2 1 50. . . .J kg m s m2= ( )( ) ( )µ πk

Thus,

µk = 0 152.

(c) After N revolutions, the object comes to rest and K f = 0.

Thus,

∆ ∆E K K mi iint = − = − + =01

22v

or

µ πk img N r m21

22( )[ ] = v

This gives

Nm

mg ri

k

=( )

= ( )( )

12

2 12

2

2

8 00

0 152 9 80

vµ π

.

. .

m s

m ss mrev2( ) ( )

=2 1 50

2 28π .

.

P8.43 (a) W K∑ = ∆ : W Ws g+ = 0

1

20 90 60 0

1

21 40 10

2

3

kx mg xi − + +( ) =

×( ) ×

∆ cos

.

° °

N m 00 100 0 200 9 80 60 0 0

4 1

2. . . sin .

.

( ) − ( )( )( ) =

=

° ∆

∆

x

x 22 m




(b) W K E∑ = +∆ ∆ int: W W Es g+ − =∆ int 0

1

2150 60 0

1

21 40 10

2

3

kx mg x mg xi k+ − =

×

∆ ∆cos cos

.

° °µ

N m( ) × ( ) − ( )( )( )

−

0 100 0 200 9 80 60 0

0

2. . . sin . ° ∆x

.. . . cos .

.

200 9 80 0 400 60 0 0

3 35

( )( )( )( ) =

=

° ∆

∆

x

x m

P8.44 P ∆ ∆∆

t W Km

= = =( )v2

2

The density is

ρ = =∆ ∆

∆m m

A xvol

Substituting this into the fi rst equation and solving for P ,

since ∆∆

x

t= v, for a constant speed, we get

P = ρAv3

2 Also, since P = Fv,

FA= ρ v2

2

Our model predicts the same proportionalities as the empirical equation, and gives D = 1 for the drag coeffi cient. Air actually slips around the moving object, instead of accumulating in front of it. For this reason, the drag coeffi cient is not necessarily unity. It is typically less than one for a streamlined object and can be greater than one if the airfl ow around the object is complicated.

P8.45 P = 1

22 3D rρπ v

(a) Pa

= ( ) ( ) ( ) = ×1

21 1 20 1 5 8 2 17 10

2 3 3. . .kg m m m s3 π WW

(b)

PP

b

a

b

a

= =⎛⎝⎜

⎞⎠⎟

= =v

v

3

3

3

324

83 27

m s

m s

Pb

= ×( )= ×27 2 17 10 5 86 103 4. .W W

*P8.46 (a) U mgy y yg = = ( )( ) = ( )64 9 8 627kg m s N2.

(b) At the original height and at all heights above 65 m − 25.8 m = 39.2 m, the cord is

unstretched and Us = 0 . Below 39.2 m, the cord extension x is given by x = 39.2 m − y,

so the elastic energy is U kx ys = = ( ) −( )1

2

1

281 39 22 2

N m m. .

(c) For y > 39 2. m, U U yg s+ = ( )627 N

For y ≤ 39 2. m,

U U y y yg s+ = ( ) + − ( ) +(627 40 5 1 537 78 4 2N N m m m2. . ))= ( ) − ( ) +40 5 2 550 62 2002. N m N Jy y

FIG. P8.44



194 Chapter 8

(d)

FIG. P8.46(d)

(e) At minimum height, the jumper has zero kinetic energy and the same total energy as at his starting point. K U K Ui i f f+ = + becomes

]

627 40 5 2 550 62 2002N 65 m N m N J( ) = ( ) − ( ) +. y yf f

00 40 5 2 550 21 500

10 0

2= − +

=

.

.

y y

y

f f

f m the root 522.9 m is unphysical.[ ]

(f ) The total potential energy has a minimum, representing a stable equilibrium position.

To fi nd it, we require dU

dy= 0:

d

dyy y y

y

40 5 2 550 62 200 0 81 2 550

31 5

2.

.

− +( ) = = −

= m

(g) Maximum kinetic energy occurs at minimum potential energy. Between the takeoff point and this location, we have K U K Ui i f f+ = +

0 40 8001

264 40 5 31 5 2 550 312 2+ = ( ) + ( ) −J kg vmax . . ..

max

5 62 200

2 40 800 22 200

6424

1 2

( ) +

=−( )⎛

⎝⎜⎞⎠⎟

=v ..1 m s

P8.47 (a) So long as the spring force is greater than the friction force, the block will be gaining speed. The block slows down when the friction force becomes the greater. It has maxi-mum speed when − − = =kx f maa k 0.

− ×( ) − =1 0 10 4 0 03. .N m Nxa x = − × −4 0 10 3. m

(b) By the same logic,

− ×( ) −1 0 10 10 03. .N m N 0xb = x = − × −1 0 10 2. m0

0

FIG. P8.47



P8.48 (a) The suggested equation P ∆t = bwd implies all of the following cases:

(1) P ∆t bw

d=⎛⎝⎜

⎞⎠⎟

( )2

2 (2) P ∆tb

wd

2 2

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

(3) P ∆tbw

d

2 2

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

and (4) P2 2

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

∆t bw

d

These are all of the proportionalities Aristotle lists.

(b) For one example, consider a horizontal force F pushing an object of weight w at constant velocity across a horizontal fl oor with which the object has coeffi cient of friction µ

k.

� �F a∑ = m implies that:

+ − =n w 0 and F nk− =µ 0

so that F wk= µ

As the object moves a distance d, the agent exerting the force does work

W Fd Fd wdk= = =cos cosθ µ0° and puts out power P = Wt∆

This yields the equation P ∆t wdk

= µ which represents Aristotle’s theory with b k= µ .

Our theory is more general than Aristotle’s. Ours can also describe accelerated motion.

P8.49 v = =100 27 8km h m s.

The retarding force due to air resistance is

R D A= = ( )( )( )1

2

1

20 330 1 20 2 50 27 82ρ v . . . .kg m m3 2 m s N( ) =2

382

Comparing the energy of the car at two points along the hill,

K U E K Ui gi f gf+ + = +∆

or

K U W R s K Ui gi e f gf+ + − ( ) = +∆ ∆

where ∆We is the work input from the engine. Thus,

∆ ∆W R s K K U Ue f i gf gi= ( ) + −( ) + −( ) Recognizing that K Kf i= and dividing by the travel time ∆t gives the required power input from

the engine as

P =⎛

⎝⎜⎞

⎠⎟=

⎛⎝⎜

⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟

= +∆∆

∆∆

∆∆

W

tR

s

tmg

y

tR me v ggvsinθ

. .P = ( )( )+( )382 27 8 1500 9 80N m s kg m s2(( )( )= =

27 8

33 4 44 8

.

. .

m s

kW hpP sin 3.20º

Ffk = µ k n

n

w

d

v = constant

FIG. P8.48


196 Chapter 8

P8.50 (a) U mgRA = = ( )( )( ) =0 200 9 80 0 300 0 588. . . .kg m s m2 JJ

(b) K U K UA A B B+ = +

K K U U mgRB A A B= + − = = 0 588. J

(c) vBBK

m= =

( )=

2 2 0 588

0 2002 42

.

..

J

kgm s

(d) U mghC C= = ( )( )( ) =0 200 9 80 0 200 0 392. . . .kg m s m2 J

K K U U mg h h

K

C A A C A C

C

= + − = −( )= ( )(0 200 9 80. .kg m s2 )) −( ) =0 300 0 200 0 196. . .m J

P8.51 (a) K mB B= = ( )( ) =1

2

1

20 200 1 50 0 2252 2v . . .kg m s J

(b) ∆ ∆ ∆E K U K K U U

K mg h h

B A B A

B B A

mech = + = − + −

= + −( )= 0 225. J kg m s m

J

2+ ( )( ) −( )

=

0 200 9 80 0 0 300

0 225

. . .

. −− = −0 588 0 363. .J J

(c) No. It is possible to fi nd an effective coeffi cient of friction, but not the actual value of µ since n and f vary with position.

P8.52 m = mass of pumpkin R = radius of silo top

F ma n mg mRr r∑ = − = −⇒ cosθ v2

When the pumpkin fi rst loses contact with the surface, n = 0. Thus, at the point where it leaves the surface: v2 = Rgcosθ.

Choose Ug = 0 in the θ = 90 0. ° plane. Then applying conservation of energy for the pumpkin-Earth system between the starting point and the point where the pumpkin leaves the surface gives

K U K U

m mgR mgR

f gf i gi+ = +

+ = +1

202v cosθ

Using the result from the force analysis, this becomes

1

2mRg mgR mgRcos cosθ θ+ = , which reduces to

cosθ =2

3, and gives θ = ( ) =−cos .1 2 3 48 2°

as the angle at which the pumpkin will lose contact with the surface.

FIG. P8.50

FIG. P8.52



P8.53 k = ×2 50 104. N m, m = 25 0. kg

xA = −0 100. m, U Ug x s x= == =

0 00

(a) E K U UA gA sAmech = + + E mgx kxA Amech = + +01

22

Emech2kg m s m= ( )( ) −( )

+ ×

25 0 9 80 0 100

1

22 50

. . .

. 110 0 100

24 5 125 100

4 2N m m

J Jmech

( ) −( )

= − + =

.

.E J

(b) Since only conservative forces are involved, the total energy of the child-pogo-stick-Earth system at point C is the same as that at point A.

K U U K U UC gC sC A gA sA+ + = + + : 0 25 0 9 80 0 0 24 5 125+ ( )( ) + = − +. . .kg m s J J2 xC

xC = 0 410. m

(c) K U U K U UB gB sB A gA sA+ + = + + : 1

225 0 0 0 0 24 5 1252. .kg J J( ) + + = + −( ) +vB

vB = 2 84. m s

(d) K and v are at a maximum when a F m= =∑ 0 (i.e., when the magnitude of the upward spring force equals the magnitude of the downward gravitational force).

This occurs at x < 0 where

k x mg=

or

x =( )( )

×= × −25 0 9 8

2 50 109 80 104

3. .

..

kg m s

N mm

2

Thus,

K K= max at x = −9 80. mm

(e) K K U U U UA gA g x sA s xmax . .= + −( ) + −

=− =−9 80 9 80mm mm(( )

or

1

225 0 25 0 9 80 0 1002. . . .maxkg kg m s2( ) = ( )( ) −v mm m( ) − −( )⎡⎣ ⎤⎦0 009 8.

+ ×( ) −( ) − −( )⎡⎣ ⎤1

22 50 10 0 100 0 009 84 2 2

. . .N m m m ⎦⎦

yielding

vmax .= 2 85 m s


198 Chapter 8

P8.54 (a) Between the second and the third picture, ∆ ∆ ∆E K Umech = +

− = − +µmgd m kdi

1

2

1

22 2v

1

250 0 0 250 1 00 9 80

1

212. . . .N m kg m s2( ) + ( )( ) −d d .. .

. .

00 3 00 0

2 45 21 35

2kg m s

N

50.0

( )( ) =

=− ±[ ]

dNN m

m= 0 378.

(b) Between picture two and picture four, ∆ ∆ ∆E K Umech = +

− ( ) = −

= ( ) − ( )

f d m m i21

2

1

2

3 002

1 002

2 2

2

v v

v ..

m skg

.. .

.

45 2 0 378

2 30

N m

m s

( )( )( )

=

(c) For the motion from picture two to picture fi ve,∆ ∆ ∆E K Umech = +

− +( ) = − ( )( )

=

f D d

D

21

21 00 3 00

9 00

2 0 2

2. .

.

.

kg m s

J

550 1 00 9 802 0 378

1 08

( )( )( ) − ( )

=

. ..

.

kg m sm

m

2

P8.55 ∆ ∆E f x

E E f d

kx mgh mgd

m

f i BC

BC

mech = −− = − ⋅

− = −

=

1

22 µ

µ ggh kx

mgdBC

− =12

2

0 328.

P8.56 Let λ represent the mass of each one meter of the chain and T represent the tension in the chain at the table edge. We imagine the edge to act like a frictionless and massless pulley.

(a) For the fi ve meters on the table with motion impending,

Fy∑ = 0: + − =n g5 0λ n g= 5λ

f n g gs s≤ = ( ) =µ λ λ0 6 5 3.

Fx∑ = 0: + − =T fs 0 T fs=

T g≤ 3λ

The maximum value is barely enough to support the hanging segment according to

Fy∑ = 0: + − =T g3 0λ T g= 3λ

so it is at this point that the chain starts to slide.

FIG. P8.54

FIG. P8.55

P

FIG. P8.56




(b) Let x represent the variable distance the chain has slipped since the start. Then length (5 − x) remains on the table, with now

Fy∑ = 0: + − −( ) =n x g5 0λ n x g= −( )5 λ

f n x g g x gk k= = −( ) = −µ λ λ λ0 4 5 2 0 4. .

Consider energies of the chain-Earth system at the initial moment when the chain starts to slip, and a fi nal moment when x = 5, when the last link goes over the brink. Measure heights above the fi nal position of the leading end of the chain. At the moment the fi nal link slips off, the center of the chain is at y

f = 4 meters.

Originally, 5 meters of chain is at height 8 m and the middle of the dangling segment is at

height 83

26 5− = . m.

K U E K Ui i f f+ + = +∆ mech : 01

21 1 2 22+ +( ) − = +⎛

⎝⎞⎠∫m gy m gy f dx m mgy

i k

i

f

f

v

5 8 3 6 5 2 0 41

28

0

52λ λ λ λ λg g g x g dx( ) + ( ) − −( ) = ( ) +∫. . v 88 4

40 0 19 5 2 00 0 4000

5

0

5

λg

g g g dx g x dx

( )

+ − + =∫ ∫. . . . 44 00 32 0

27 5 2 00 0 4002

4 0

2

0

52

0

5

. .

. . . .

v +

− + =

g

g gx gx

00

27 5 2 00 5 00 0 400 12 5 4 00

22

2

2

v

v. . . . . .g g g− ( ) + ( ) =

.. .

. .

..

5 4 00

22 5 9 80

4 007 42

2g =

=( )( )

=

v

vm m s

m s2

P8.57 K U K Ui f+( ) = +( )

0 30 0 9 80 0 2001

2250 0+ ( )( )( ) + ( ). . . .kg m s m N m2 2200

1

250 0 20 0 9 80

2

2

m

kg kg m s2

( )

= ( ) + ( )( ). . .v 00 200 40 0. sin .m( ) °

58 8 5 00 25 0 25 2

1 24

2. . . .

.

J J kg J

m s

+ = ( ) +

=

v

vFIG. P8.57


200 Chapter 8

P8.58 The geometry reveals D L L= +sin sinθ φ , 50 0 40 0 50. . sin sinm m= +( )° φ , φ = 28 9. °

(a) From takeoff to alighting for the Jane-Earth system

K U W K U

m mg L FD

g i g f

i

+( ) + = +( )+ −( ) + −

wind

1

212v cosθ (( ) = + −( )0 mg L cosφ

1

250 50 9 8 40 50 1102kg kg m s m2vi + ( ) −( ) −. cos ° NN m

kg m s m2

50

50 9 8 40 28 9

1

250

( )

= ( ) −( ). cos . °

kkg J J Jv

v

i

i

2 4 3 41 26 10 5 5 10 1 72 10

2

− × − × = − ×

=

. . .

9947

506 15

J

kgm s

( )= .

(b) For the swing back

1

21 0

1

2130

2m mg L FD mg Liv + −( ) + +( ) = + −( )cos cosφ θ

kkg kg m s m N2vi2 130 9 8 40 28 9 110+ ( ) −( ) +. cos . ° 550

130 40 50

1

2130

m

kg 9.8 m s m

k

2

( )

= ( ) −( )cos °

gg J J Jv

v

i

i

2 4 44 46 10 5 500 3 28 10

2 6 340

− × + = − ×

=

. .

J

kgm s

( ) =130

9 87.

P8.59 (a) Initial compression of spring: 1

2

1

22 2kx m= v

1

2450

1

20 500 12 0

2 2N m kg m s

There

( )( ) = ( )( )∆x . .

ffore, m∆x = 0 400.

(b) Speed of block at top of track: ∆ ∆E f xmech = −

mgh m mgh m f RT T B B+⎛⎝

⎞⎠ − +⎛

⎝⎞⎠ = − ( )1

2

1

2

0 500

2 2v v π

. kg m s m kg2( )( )( ) + ( ) −9 80 2 001

20 500

1

22. . . vT 00 500 12 0

7 00 1 00

2. .

. .

kg m s

N m

( )( )= −( )( )π (( )

=

∴ =

0 250 4 21

4 10

2. .

.

v

v

T

T m s


VT

FIG. P8.59



(c) Does block fall off at or before top of track? Block falls if a gc <

aRcT= =

( )=

v2 24 10

1 0016 8

.

.. m s2

Therefore a gc > and the block stays on the track .

*P8.60 (a) Take the original point where the ball is released and the fi nal point where its upward swing stops at height H and horizontal displacement

x L L H LH H= − −( ) = −2 2 22

Since the wind force is purely horizontal, it does work

W d F dx F LH Hwind = ⋅ = = −∫ ∫� �F s 2 2

The work-energy theorem can be written:

K U W K Ui gi f g f+ + = +wind , or

0 0 2 02+ + − = +F LH H mgH giving

F LH F H m g H2 2 2 2 2 22 − =

Here the solution H = 0 represents the lower turning point of the ball’s oscillation, and the

upper limit is at F L F m g H2 2 2 22( ) = +( ) .Solving for H yields

HLF

F m g

L

mg F=

+=

+ ( )=

+2 2

1

2 0 82

2 2 2 2

( . m)

1 (0.3 kg)22 2 2(9.8 m/s

m

1 8.64 N)

.2 2 2

1 6

� �F F=

+

(b) H = + =−1.6 m 1 8.64 1 0.166 m1[ ]

(c) H = + =−1.6 m 1 8.64 1 1.47 m100[ ]

(d) As F → 0, H → 0 as is reasonable.

(e) As F → ∞, H → 1.60 m, which would be hard to approach experimentally.

(f ) Call θ the equilibrium angle with the vertical and T the tension in the string.

F T F

F T mg

x

y

∑∑

= ⇒ =

= ⇒ =

0

0

sin

cos

θ

θ

, and

Dividing: tanθ =F

mg

Then

cosN2

θ = mg

mg F F mg F( ) ( ) .2 2 2 2

1

1

1

1 8 64+=

+=

+� �

Therefore,

H LF

eq 2m

N= −( ) = ( ) −

+

⎛⎝⎜

⎞⎠

1 0 800 11

1 8 642cos .

.θ

�⎟⎟

FIG. P8.60



202 Chapter 8

(g) Heq

= 0.8 m[1 − (1 + 100�8.64)−1� 2] = 0 574. m

(h) As F → ∞, tan θ → ∞, θ → 90.0° cos θ → 0 and Heq → 0.800 m

A very strong wind pulls the string out horizontal, parallel to the ground.

P8.61 If the spring is just barely able to lift the lower block from the table, the spring lifts it through no noticeable distance, but exerts on the block a force equal to its weight Mg. The extension of the spring, from Fs kx= , must be Mg k. Between an initial point at release and a fi nal point when the moving block fi rst comes to rest, we have

K U U K U Ui gi si f gf sf+ + = + + : 04 1

2

40

1

2

2

+ −⎛⎝

⎞⎠ + ⎛

⎝⎞⎠ = + ⎛

⎝⎞⎠ +mg

mg

kk

mg

kmg

Mg

kk

Mgg

k⎛⎝

⎞⎠

2

− + = +

= +

+

4 8

2

42

2

2 2 2 2 2 2 2

22

2

m g

k

m g

k

mMg

k

M g

k

m mMM

MmM −− =

=− ± − ( ) −( )

( ) = − ±

4 0

4 4

29

2

2 12

2

12

2

m

Mm m m

m m

Only a positive mass is physical, so we take M m m= −( ) =3 1 2 .

P8.62 (a) Energy is conserved in the swing of the pendulum, and the stationary peg does no work. So the ball’s speed does not change when the string hits or leaves the peg, and the ball swings equally high on both sides.

(b) Relative to the point of suspension,

Ui = 0, U mg d L df = − − −( )[ ] From this we fi nd that

− −( ) + =mg d L m21

202v

Also for centripetal motion,

mgm

R=

v2

where R L d= −

Upon solving, we get dL

=3

5.

P8.63 Applying Newton’s second law at the bottom (b) and top (t) of the circle gives

T mgm

Rbb− =

v2

and − − = −T mgm

Rttv2

Adding these gives

T T mgm

Rb tb t= + +

−( )2

2 2v v

Also, energy must be conserved and ∆ ∆U K+ = 0

So, m

mgRb tv v2 2

20 2 0

−( )+ −( ) = and

m

Rmgb tv v2 2

4−( )

=

Substituting into the above equation gives T T mgb t= + 6 .

θL

d

Peg

FIG. P8.62

mg

v t

Ttmg

Tb

vb

FIG. P8.63



v

FIG. P8.64

P8.64 (a) At the top of the loop the car and riders are in free fall:

F may y∑ = : mgm

Rdown down=

v2

v = Rg

Energy of the car-riders-Earth system is conserved between release and top of loop:

K U K Ui gi f gf+ = + : 01

222+ = + ( )mgh m mg Rv

gh Rg g R

h R

= + ( )

=

1

22

2 50.

(b) Let h now represent the height ≥ 2.5 R of the release point. At the bottom of the loop we have

mgh m b=1

22v or vb gh2 2=

F may y∑ = : n mgm

Rbb− = ( )v2

up

n mgm gh

Rb = +( )2

At the top of the loop,

mgh m mg R

gh gR

t

t

= + ( )

= −

1

22

2 4

2

2

v

v

F may y∑ = : − − = −n mgm

Rttv2

n mgm

Rgh gR

nm gh

Rmg

t

t

= − + −( )

=( )

−

2 4

25

Then the normal force at the bottom is larger by

n n mgm gh

R

m gh

Rmg mgb t− = +

( )−

( )+ =

2 25 6

P8.65 (a) Conservation of energy for the sled-rider-Earth system, between A and C:

K U K U

m m

i gi f gf+ = +

( ) + ( )1

22 5 9 80 9 76

2. . .m s m s m2 (( )

= +

= ( ) + ( )(

1

20

2 5 2 9 80 9 76

2

2

mv

v

C

C2m s m s m. . . ))

= 14 1. m scontinued on next page

FIG. P8.65(a)


204 Chapter 8

(b) Incorporating the loss of mechanical energy during the portion of the motion in the water, we have, for the entire motion between A and D (the rider’s stopping point),

K U f x K Ui gi k f gf+ − = +∆ :

1

280 2 5 80 9 80 9 76

2kg m s kg m s m2( )( ) + ( )( )(. . . )) − = +f xk ∆ 0 0

− = − ×f xk ∆ 7 90 103. J

(c) The water exerts a frictional force

fxk =

×=

× ⋅=

7 90 10 7 90 10158

3 3. .J N m

50 mN

∆ and also a normal force of

n mg= = ( )( ) =80 9 80 784kg m s N2.

The magnitude of the water force is

158 784 8002 2N N N( ) + ( ) =

(d) The angle of the slide is

θ = =−sin.

.1 9 7610 4

m

54.3 m°

For forces perpendicular to the track at B,

F may y∑ = : n mgB − =cosθ 0

nB = ( )( ) =80 0 9 80 10 4 771. . cos .kg m s N2 °

(e) F may y∑ = : + − =n mgm

rCC2v

nC2kg m s

kg m s

= ( )( )

+( )( )80 0 9 80

80 0 14 1

2

2

. .

. .

00

1 57 103

m

N upCn = ×.

The rider pays for the thrills of a giddy height at A, and a high speed and tremendous splash at C. As a bonus, he gets the quick change in direction and magnitude among the forces we found in parts (d), (e), and (c).

*P8.66 (a) As at the end of the process analyzed in Example 8.8, we begin with a 0.800-kg block at rest on the end of a spring with stiffness constant 50.0 N/m, compressed 0.0924 m. The energy in the spring is (1/ 2)(50 N/m)(0.0924 m)2 = 0.214 J. To push the block backto the unstressed spring position would require work against friction of magnitude3.92 N(0.0924 m) = 0.362 J. Because 0.214 J is less than 0.362 J, the spring cannot push the object back to x = 0.

FIG. P8.65(d)


FIG. P8.65(e)



(b) The block approaches the spring with energy (1� 2)mv2 = (1� 2)(0.8 kg)(1.2 m �s)2 = 0.576 J. It travels against friction by equal distances in compressing the spring and in being pushed back out, so it must lose one-half of this energy in its motion to the right and the rest in its motion to the left. The spring must possess one-half of this energy at its maximum com-pression:

(0.576 J)� 2 = (1� 2) (50 N�m)x2

so

x = 0.107 m

For the compression process we have the continuity equation for energy

0.576 J + µk 7.84 N (0.107 m) cos 180° = 0.288 J

so

µk = 0.288 J�0.841 J = 0.342

As a check, the decompression process is described by

0.288 J + µk7.84 N (0.107 m) cos 180° = 0

which gives the same answer for the coeffi cient of friction.

ANSWERS TO EVEN PROBLEMS

P8.2 (a) 1 11 109. × J (b) 0.2

P8.4 (a) vB = 5 94. m s; vC = 7 67. m s (b) 147 J

P8.6 (a) see the solution (b) 60.0°

P8.8 (a) 2 1 2

1 2

m m gh

m m

−( )+( )

(b) 2 1

1 2

m h

m m+

P8.10 (a) 18.5 km, 51.0 km (b) 10.0 MJ

P8.12 8

15

1 2gh⎛⎝

⎞⎠

P8.14 (a) 0.791 m �s (b) 0.531 m �s

P8.16 (i) (a), (b), (c), (f ) (ii) (g), (i), ( j) (iii) (d) (iv) (e) cannot be true because the friction force is proportional to µ

k and not µ

k2. And (k) cannot be true because the presence of friction will reduce

the speed compared to the µk = 0 case. (v) Expression (h) is correct if the spring force is strong

enough to produce motion against static friction and if the spring energy is large enough to make the block slide the full distance d. (vi) The expression gives an imaginary answer because the spring does not contain enough energy in th is case to make the block slide the full distance d.

P8.18 (a) U f = 22 0. J; E = 40 0. J (b) Yes. The total mechanical energy changes.

P8.20 26.5 m �s


206 Chapter 8

P8.22 (a) 24 5. m s (b) Yes; his landing speed is too high to be safe. (c) 206 m (d) Not realistic. Air drag depends strongly on speed.

P8.24 (a) r = 1 5. mm and 3.2 mm, stable; 2.3 mm unstable; r → ∞ neutral (b) − ≤ <5 6 1. J JE (c) 0 6 3 6. .mm mm≤ ≤r (d) 2.6 J (e) 1.5 mm (f ) 4 J

P8.26 168 J

P8.28 8.01 W

P8.30 (a) 1.24 kW (b) 20.9%

P8.32 (a) 5.91 kW (b) 11.1 kW

P8.34 194 m

P8.36 No. (a) 582 (b) 90 5 0 121. .W hp=

P8.38 (a) yes (b) 2.49 m�s (c) No, but mechanical energy is conserved for the 3.50-kg block in its projectile motion with the Earth. (d) 5.45 m�s (e) 1.23 m (f ) no (g) 18.6 N to the left (h) A little push is required. The speeds are still accurate.

P8.40 (a) x = 0.403 m or −0.357 m (b) From a perch at a height of 2.80 m above the top of a pile of mattresses, a 46.0-kg child jumps straight upward at 2.40 m �s. The mattresses behave as a linear spring with force constant 19.4 kN�m. Find the maximum amount by which the mattresses are compressed when the child lands on them. Physical meaning of the answer: The positive value of x represents the maximum spring compression. The negative value represents the extension of the equivalent spring if the child sticks to the top of the mattress pile as the child rebounds upward without friction.

P8.42 (a) 5.60 J (b) 0.152 (c) 2.28 rev

P8.44 See the solution. Our model predicts the same proportionalities as the empirical equation, and gives D = 1 for the drag coeffi cient. Air actually slips around the moving object, instead of accumulating in front of it. For this reason, the drag coeffi cient is not necessarily unity. It is typically less than one for a streamlined object and can be greater than one if the airfl ow around the object is complicated.

P8.46 (a) 627 N( ) y (b) Us = 0 for y > 39 2. m and U ys = ( ) −( )1

281 39 2

2N m m. for y ≤ 39 2. m

(c) U U yg s+ = ( )627 N , for y > 39 2. m and U U yg s+ = ( )40 5 2. N m −( ) +2 550 62 200N Jy for y ≤ 39 2. m (d) see the solution (e) 10.0 m (f ) yes: stable equilibrium at 31.5 m (g) 24.1 m�s

P8.48 (a) see the solution (b) For a block of weight w pushed over a rough horizontal surface at constant velocity, b = µ

k. For a load pulled vertically upward at constant velocity, b = 1.

P8.50 (a) 0.588 J (b) 0.588 J (c) 2 42. m s (d) UC = 0 392. J, KC = 0 196. J

P8.52 48.2°

P8.54 (a) 0.378 m (b) 2 30. m s (c) 1.08 m

P8.56 (a) see the solution (b) 7 42. m s



P8.58 (a) 6.15 m �s (b) 9.87 m �s

P8.60 (a) H = 1.6 m(1 + 8.64 N2� F2 )−1 (b) 0.166 m (c) 1.47 m (d) H → 0 proportionally to F 2

(e) H approaches 1.60 m (f ) Heq

= 0.8 m[1 − (F 2�8.64 N2 + 1)−1�2] (g) 0.574 m (h) 0.800 m

P8.62 see the solution

P8.64 (a) 2.5 R (b) see the solution

P8.66 (a) see the solution (b) 0.342


sm chapter8

Education