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Conservation of Energy
CHAPTER OUTLINE 8.1 The Nonisolated System— Conservation of Energy 8.2 The Isolated System 8.3 Situations Involving Kinetic Friction 8.4 Changes in Mechanical Energy for Nonconservative Forces 8.5 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 field.
*Q8.3 (i) answer b. Kinetic energy is proportional to mass.
(ii) answer c. The slide is frictionless, so v = (2 gh )^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 floor where we choose y = 0.
(b) yes: a picture on the classroom wall high above the floor.
(c) yes: an eraser hurtling across the room.
(d) yes: the block stationary on the floor.
*Q8.5 answer (d). The energy is internal energy. Energy is never “used up.” The ball finally 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
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 ( U (^) s). 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 (12) mv^2 = μ kmgd so d = v^2 2 μ kg. The quantity v^2 μ k controls the skidding distance. In the cases quoted respectively, this quantity has the numerical value (a) 5 (b) 2.5 (c) 1. (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 flat bed of a truck exerts a static friction force to start a pumpkin moving forward as it slowly starts up.
*Q8.14 (a) A campfire 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% efficient. 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 efficiency 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 efficiency.
(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% efficiency 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 efficiency 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 efficiency, the pressure of light from a newborn star pushes away gas and dust in the nebula surrounding it.
176 Chapter 8
FIG. Q8.
178 Chapter 8
P8.3 U i + K i = U f + Kf : mgh + 0 = mg ( 2 R ) + m
v^2
g 3 50 R 2 g R^1 2
(. ) = ( ) + v^2
v = 3 00. gR
F m ∑ (^) R = v
2 : n mg m R
2
n m R
g m gR R
g mg
n
⎥ =^ −
⎣⎢^
v^2 3 00 (^) 2 00
( × )( )
− (^) kg m s
N downwar
2
dd
P8.4 (a) ( ∆ K ) A (^) → B = (^) ∑ W = W (^) g = mg ∆ h = mg ( 5 00. −3 20. )
1 2
m^2 m^2 m B A
B
v v
v
. m s
Similarly, v C = vA^2 + 2 g ( 5 00. −2 00. ) = 7 67. m s
(b) W g A → C = mg ( 3 00. m ) = 147 J
P8.5 From conservation of energy for the block-spring-Earth system,
U (^) gt = Usi or 0 250 9 80
(. kg )(. m s (^2) ) h = ⎛⎝ ⎞⎠ (5 000 N m)( 0 100. 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 cos θ = m v^2 or
F = mg cos θ + mv
2 Apply conservation of mechanical energy of the performer- Earth system as the performer moves between the starting point and any later point:
mg (^) ( − cos θ i ) = mg ( −cosθ ) +^1 m 2
v^2
Solve for mv
2
and substitute into the force equation to obtain F = mg (^) ( 3 cos θ − 2 cosθ i (^) ).
FIG. P8.
5.00 m
A
B
C
2.00 m
3.20 m
FIG. P8.
FIG. P8.
FIG. P8.
continued on next page
Conservation of Energy 179
(b) At the bottom of the swing, θ = 0° so
F mg F mg mg
i i
= (^) ( − ) = = (^) ( − )
cos cos
θ θ which gives
θ i = 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 00
(. kg ) g (. m ) = (. kg ) g (. m) + (5 0. 00 +3 00. ) v^2
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 2
max.
v mg y g y y y
∆ m m ++ ∆ y = 5 00. m
P8.8 We assume m 1 (^) > m 2
(a) (^) m gh 1^1 m (^) 1 m (^) 2 2 m gh 2 2 = ( + ) v +
v = ( − ) ( + )
1 2
m m gh m m
(b) Since m 2 has kinetic energy
m v^2 , it will rise an additional height ∆ h determined from
m g 2 h m 2 2
∆ = v
or from (a),
∆ h g
m m h m m
( − ) ( + )
v^21 (^21 )
The total height m 2 reaches is h h m h m m
1 2
FIG. P8.
Conservation of Energy 181
(b) k = 3 33 mg 25 8
. m
x (^) max = x (^) f = 55 0. m −25 8. m =29 2. m
∑^ F^ =^ ma +^ kx^ max−^ mg^ = ma
3 33 25 8
mg (^) mg ma
a g
m
m
m s 2
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 final point to be when B has moved up by h 3
and has speed v A 2
Then A has moved down 2 3
h (^) and has speed v A :
K K U K K U
m m
A B g (^) i A B g (^) f
A 2 A
( +^ + ) =^ ( +^ + )
v (^) ⎝⎝ v ⎞⎠ + −
=
2 3
mgh mg h
mgh (^) m
gh
v
v
A 2
A
Section 8.3 Situations Involving Kinetic Friction
P8.13 (^) ∑ Fy = may : n − 392 N= 0
n f (^) k kn
N
μ. N N
(a) W F = F ∆ r cos θ = ( 130 )( 5 00. ) cos 0 ° = 650 J
(b) ∆ E int = f k ∆ x = ( 118 )( 5 00. ) = 588 J
(c) Wn = n ∆ r cos θ = ( 392 )( 5 00. ) cos 90 °= 0
(d) W g = mg ∆ r cos θ = ( 392 )( 5 00. ) cos(− 90 °) = 0
(e) ∆ K = K (^) f − K (^) i = (^) ∑ W (^) other −∆ E int 1 2 m v^2 f − 0 = 650 J − 588 J + 0 + 0 = 62 0. J
(f) v (^) f f
K
m
J
kg m s
FIG. P8.
182 Chapter 8
P8.14 (a) W (^) s = kx (^) i − kxf
= ( ) (^) ( × −) − =
2 2
. 2 2. 5 5 J
W (^) s = m (^) f − m (^) i = m f −
v^2^ v^2^ v^20
so
v (^) f
W
m
(^2) ∑
m s. m s
(b)
m^2 f x W m^2 vi − (^) k ∆ + (^) s = vf
− (. )(. )(. )(. ) J +. J = mvv
v
v
f
f
f
2
J kg
m
= ( )
ss = 0 531. m s
P8.15 (a) W g = mg cos ( 90 0°. + θ)
W (^) g = ( )( (^) )( ) = −
. kg. m s 2. m cos ° JJ
(b) f (^) k = μ k (^) n =μ k mg cosθ ∆ ∆
E f mg E
int k k int m
μ cosθ 5 00. 0 400. 10 0. 0 9 80 20 0 184
. cos. ° J
(c) W F = F = ( 100 )( 5 00. ) = 500 J
(d) ∆ K = (^) ∑ W (^) other − ∆ E (^) int = W (^) F + W (^) g − ∆ E int = 148 J
(e) ∆ K = m (^) f − m i
v^2 v^2
v (^) f K vi m
= (^
+ = (^
.. 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 (12) kd^2 = (12) mv^2 and v = ( 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.
continued on next page
FIG. P8.
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 cos 0°= F ( π R )
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 (^) mech = m (^) f − m (^) i + mgy (^) f − mgyi
v^2 v^2
becomes 1 2
m^2 m^2 mgy F R
v f = vi + i + ( π )
or
v (^) f vi gyi
F R
m
π 2
...
v (^) f = 26 5. m s
P8.21 (a) ∆ K = m ( f − i ) = − m i = −
v^2^ v^2 v^2 160 J
(b) ∆ U = mg ( 3 00. m) sin 30 0. °= 73 5. J
(c) The mechanical energy converted due to friction is 86.5 J
f = =
J
3.00 m
N
(d) f^ =^ μ k^ n^ =^ μ k^ mg cos^ 30 0.^ °^ =28 8. N
μ k = ( )( )
. cos.
N
5.00 kg m s 2 °
P8.22 Consider the whole motion: K^ i +^ U^ i +^ ∆ E^^ mech=^ K^ f + Uf
(a) 0
1 1 2 2
( )
mgyi f ∆ x f ∆ x mvf
. kg (^) (. m s (^22) ) 1 000 m − ( 50 0. N )( 800 m ) − ( 3 600 N )( 200 m) =
. kg^2
J J J
( )
− − =
v (^) f
8 80 0
2 24 000 80 0
.^2
kg
J kg
m s
( )
( )
v
v
f
f
(b) Yes. This is too fast for safety.
(c) Now in the same energy equation as in part (a), ∆ x 2 is unknown, and ∆ x (^) 1 = 1 000m −∆ x 2 :
784 000 50 0 1 000 3 600
J − (. N ) (^) ( m −∆ x (^) 2 ) − ( N) ∆ x 2 = 22 80 0 5 00
784 000 50 000 3 550
. kg. m s^2
J J
( )( )
− − NN J J 3 550 N
m
( ) =
= =
x
x
2
2
FIG. P8.
continued on next page
Conservation of Energy 185
(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 ) + i ∆ E mech = ( K + U ) f :
2 2
2
( ) (^) ( × − )
kx f ∆ x mv
. N m. m (^22 2 ) 3 20 10 0 150 1 2
− (^) (. × −^ N)(. m ) = (^) (5 30. × 10 − kg) v^22 3 3
v = 1 40 ( × ) ×
− −
J
kg m s
(b) When the spring force just equals the friction force, the ball will stop speeding up. Here F s = kx ; the spring is compressed by 3 20 10 0 400
×
− N
8.00 N m
cm
and the ball has moved 5 00. cm − 0 400. cm = 4 60. cm from the start.
(c) Between start and maximum speed points,w 1 2
2 2 2
2 2
kx (^) i − f x = m + kxf
( × −) −
∆ v
.. ... ..
2 2
3 2
( × ) ( × )
= (^) ( × ) +
− −
− (^) v 0 00 4 00 10
1 79
.^3
( × )
=
−
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. J ≤ E < 1 J.
(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. mm ≤ r ≤3 6. mm.
(d) K + U = E. Thus, K max = E − U min = − 3 0. J − −( 5 6. J) = 2 6. J.
(e) Kinetic energy is a maximum when the potential energy is a minimum, at (^) r = 1 5. mm.
(f) − 3 J + W = 1 J. Hence, the binding energy is W = 4 J.
Conservation of Energy 187
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^ μ k^ mgx L
= ma. Then
a gx L =μ k^ opposite to the motion..
(b) In an incremental bit of forward motion dx , the kinetic energy converted into internal energy is (^) f dx mgx L k^ = k dx
μ (^). The whole energy converted is
1 2 2 2
2 0
2
0
m mgx L
dx mg L
x mgL
gL
k
L 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 t
f ∆ ∆ ∆
v^2 2
( )
s
. W
P8.29 Power = W t
P = = ( )( ) mgh = t
N m s
W
*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 (^) i + U (^) gi + W (^) pump = K (^) f + Ugf , subtract out the K terms, and choose U (^) gi = 0 at the bottom of the sump, to obtain W (^) pump = mgy (^) f. Now we differentiate through with respect to time:
Ppump
kg m 3 L d
= ×
m t
gy V t f^ ρ gyf
(^1050) ((1 89. (^10 6) )
m 1 9 80 L
d 86 400 s
m s
3 2
( )
= ×
m
W
(b) (^) efficien y useful output work total input wo c = rrk
useful output work total input work
t t mmechanical output power input electric powerr
kW 5.90 kW
The remaining power, (^) 5 90. − 1 24. kW =4 66. 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 ms in 15.0 s. The output work of the engine is equal to its final kinetic energy,
1 2
(1 300 kg )(24 6. m s )^2 = 390 kJ
with power (^) P = 390 000^ J 10 4 15.0 s
~ W around 30 horsepower.
P8.32 (a) The distance moved upward in the first 3.00 s is
∆ y = t =
⎣⎢^
v ( ) =
m s s m
The motor and the earth’s gravity do work on the elevator car: 1 2
m^2 W mg y m^2
W
vi + + = vf
=
motor
motor
∆ cos °
( 00 kg )(1 75. m s ) −^2 0 + ( 650 kg ) g ( 2 63. m) = 1 77. × 110 4 J
Also, W =P t so P = =
×
= × =
W
t
J
3.00 s W hp.
(b) When moving upward at constant speed ( v =1 75. m s )the applied force equals the weight = ( 650 kg )( 9 80. m s (^2) ) = 6 37. × 10 3 N. Therefore,
P = Fv = (^) ( 6 37. × 10 3 N (^) )(1 75. m s (^) ) = 1 11. × 10 4 W =14 9. hhp
P8.33 energy = power × time For the 28.0 W bulb: Energy used = ( 28 0. W )^ ( 1 00. × 10 4 h) = 280 kilowattt hrs⋅ total cost = $ 17 00. + ( 280 kWh )( $ .0 080 kWh) =$ 39 4. 40
For the 100 W bulb: Energy used = ( 100 W ) (^) ( 1 00. × 10 4 h (^) ) = 1 00. × 103 killowatt hrs⋅
bulb used h
750 h bulb
= 1 00^ ×^10 =13 3
total cost = 13 3. ( $ .0 420 ) + (^) ( 1 00. × 10 3 kWh)($. 0 08 00 kWh) = $85 60. Savings with energy-efficient bulb = $ 85 60. −$ 3 39 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 ∆
∆ (( (^) ) ⋅
⎝⎜^
⎠⎟^
⎝⎜^
N
J
W s
N m J
m
190 Chapter 8
Additional Problems
P8.37 (a) ( K + U g ) A = ( K + Ug ) B
- mgy (^) A = mv^2 B + 0 vB = 2 gy (^) A = 2 9 8(. m s (^2) ) 6 3. m = 11 1. m s
(b) a c (^) r
( )
v^2 11 1^2 6 3
m s m m s 2 up
(c) (^) ∑ Fy = may + n (^) B − mg = mac n (^) B = 76 kg (^) ( 9 8. m s 2 +19 6. m s (^2) ) = 2 23. × 10 3 N up
(d) We compute the amount of chemical energy converted into mechanical energy as
W = F ∆ r cos θ = 2 23. × 10 3 N ( 0 450. m ) cos 0 ° = 1 01. × 103 J
(e) ( K + U g + U chemical ) B = ( K + Ug ) D
m^2 3 m^2 mg y y vB + +. × J= vD + ( (^) D − B )
kg 111 1 1 01 10
(. m s ) +^2. × 3 J = 76 kg vD^2 + 76 kg9 8. mm s m
J J kg
( 2 )
( ×^ −^ × )
3 3
vD == 5 14. m s
(f) ( K + U g ) D = ( K + Ug ) E where E is the apex of his motion
m v^2 D + 0 = 0 + mg y ( (^) E − yD ) y y E D g − = D = (^ ) ( )
v^2 2
m s m s 2 m (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
− = + + −
v
2
. m. m s (^) (. mm s (^2) )
− − =
=
t
t t
t
2
2 2
. 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 floor m 2 gy = (12) ( m 1 + m 2 ) v^2 v = [2 m 2 gy ( m 1 + m 2 )] 1 ^2 = [2(1.90 kg)(9.8 ms 2 )0.9 m5.4 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.
continued on next page
Conservation of Energy 191
(d) For the 3.5-kg block from when the string goes slack until just before the block hits the fl oor (12) ( m 2 ) v^2 + m 2 gy = (12) ( m 2 ) vd^2 vd = [2 gy + v^2 ] 1 ^2 = [2(9.8 ms 2 )1.2 m + (2.49 ms) 2 ] 1 ^2 = 5.45 m /s
(e) The 3.5-kg block takes this time in flight to the floor: from y = (1 2) gt^2 we have t = [2(1.2)9.8]^1 ^2 = 0.495 s. Its horizontal component of displacement at impact is then x = vd t = (2.49 ms)(0.495 s) = 1.23 m.
(f) No. With the hanging block firmly 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 μ sn = (0.56)(3.5 kg)(9.8 ms 2 ) = 19.2 N. The hanging block tends to produce string tension (1.9 kg)(9.8 ms 2 ) = 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 + 2 00. t^3
Therefore, v
v
= = ( ) ( + ) =
dx dt
t
K m t
2
2 2 2
.. ( 2 2 00. + 24 0. t^2 +72 0. t^4 )J
(b) a d dt
= = ( t )
v 12 0. m s 2
F = ma = 4 00 12 0. (. t ) = ( 48 0. t )N
(c) P = Fv = ( 48 0. t (^) ) ( 1 +6 00. t^2 ) = (^) ( 48 0. t + 288 t^3 )W
(d) (^) W = (^) ∫ P dt = (^) ∫( t + t (^) ) dt = 0
2 00 3 0
2 00 48 0 288 1250
.. . J
*P8.40 (a) Simplifi ed, the equation is 0 = ( 9 700 N m ) x^2 − ( 450 8. N) x − 1 395N m⋅. Then
x b b ac a
(^2 4) ± ( ) −^2 ( ) 2
450 8. N 450 8. N 4 9 700N m ( −− ⋅ ) ( )
= ±
N m N m N N 19 400
N m
= 0 403. m or −0 357. m
(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 ms. The mattresses behave as a linear spring with force constant 19.4 kNm. 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.
Conservation of Energy 193
(b) (^) ∑ W = ∆ K +∆ E int: W (^) s + W (^) g − ∆ E (^) int = 0 1 2
2
3
kx (^) i + mg x − (^) kmg x =
×
∆ cos cos ∆
.
° μ °
( N m) × (^ ) − (^ )(^ )(^ ) −
. 2.. sin. ° ∆ x .... cos. .
x x m
P8.44 P ∆ ∆
t W K
m = = =
( ) v^2 2 The density is ρ = =
m m vol A x Substituting this into the first equation and solving for P , since
x t
= v , for a constant speed, we get
P = ρA v^3 2 Also, since P = Fv ,
F
A
ρ v^2 2
Our model predicts the same proportionalities as the empirical equation, and gives D = 1 for the drag coefficient. Air actually slips around the moving object, instead of accumulating in front of it. For this reason, the drag coefficient is not necessarily unity. It is typically less than one for a streamlined object and can be greater than one if the airflow around the object is complicated.
P8.45 P =
D ρπ r^2 v^3
(a) P a = (^1) ( ) ( ) ( ) = × 2
(^2 3 )
. kg m 3 π. m m s. WW
(b)
P P
b a
b a
⎠⎟^
v v
3 3
3 (^243) 8
m s 3 27 m s
P b = 27 2 17(. × 10 3 W (^) ) = 5 86. × 104 W
*P8.46 (a) U (^) g = mgy = ( 64 kg )( 9 8. m s (^2) ) y = ( 627 N) y
(b) At the original height and at all heights above 65 m − 25.8 m = 39.2 m, the cord is unstretched and U (^) s = 0. Below 39.2 m, the cord extension x is given by x = 39.2 m − y , so the elastic energy is U (^) s = kx = ( ) ( − y )
(^2 81) N m 39 2. m^2.
(c) For y > 39 2. m, U g + U s = ( 627 N) y
For y ≤ 39 2. m, U (^) g + U (^) s = ( 627 N ) y + 40 5. N m (^) (1 537 m 2 − ( 78 4. m) y + y^2 )) = ( 40 5. N m ) y^2 − ( 2 550 N ) y +62 200J
FIG. P8.
continued on next page
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 (^) i + U (^) i = K (^) f + Uf becomes
]
627 N 65 m( ) = ( 40 5. N m ) y^2 (^) f − ( 2 550 N ) yf +62 200J 00 40 5 2 550 21 500 10 0
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
d dy y y y
y
.^2
( − + ) = = −
= m
(g) Maximum kinetic energy occurs at minimum potential energy. Between the takeoff point and this location, we have K (^) i + U (^) i = K (^) f + Uf
0 40 800 1 2
- J = (^) ( 64 kg) v^2 max^ + 40 5 31 5. (. ) −^2 2 550 31..
max
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 (^) a − f (^) k = ma = 0.
− (^) ( 1 0. × 10 3 N m (^) ) x (^) a − 4 0. N= 0 x = − 4 0. × 10 −^3 m
(b) By the same logic, − (^) ( 1 0. × 10 3 N m (^) ) x (^) b −10 0. N = 0 x = − 1 0. × 10 − 2 m 0
FIG. P8.
