Download CHAPTER 6 WORK AND ENERGY

CHAPTER 6 WORK AND ENERGY
PROBLEMS
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1.
SSM REASONING AND SOLUTION We will assume that the tug-of-war rope remains
parallel to the ground, so that the force that moves team B is in the same direction as the
displacement. According to Equation 6.1, the work done by team A is
3
W = ( F cosθ )s = (1100 N)(cos 0 °)(2.0 m) = 2.2 × 10 J
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2.
REASONING The work done by a constant force is given by Equation 6.1 as
W = ( F cos θ ) s , where F is the magnitude of the force, s is the magnitude of the
F
displacement, and θ is the angle between the force and the displacement. When the
amplifier is being lowered at a constant velocity, it is in equilibrium so the net force Amplifier
in the vertical direction is zero. Thus, the magnitude of the force F exerted by the
student is equal to the magnitude of the amplifier’s weight mg. Since the amplifier is
moving down, the angle between the force F and the displacement s is θ = 180°.
mg
SOLUTION
a. The work done by the student in lowering the amplifier is
W = (F cos θ)s = (mg cos θ)s
= (10.5 kg)(9.80 m/s2)(cos 180°)(1.82 m − 0.75 m) = −1.10 × 102 J
b. The work done by the gravitational force (the weight mg of the amplifier) is
W = (mg cos θ )s
= (10.5 kg)(9.80 m/s2)(cos 0°)(1.82 m − 0.75 m) = +1.10 × 102 J
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3.
REASONING AND SOLUTION According to Equation 6.1, W = Fs cos θ, the work is
a.
W = (94.0 N)(35.0 m) cos 25.0° = 2980 J
b.
W = (94.0 N)(35.0 m) cos 0° = 3290 J
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s
Chapter 6 Problems
4.
REASONING AND SOLUTION
207
Each locomotive does work
W = Ts cos θ = (5.00 × 103 N)(2.00 × 103 m) cos 20.0° = 9.40 × 106 J
The net work is then
WT = 2W = 1.88 × 10 7 J
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5.
SSM REASONING AND SOLUTION Solving Equation 6.1 for the angle θ, we obtain
W 
1.10 × 103 J 
−1 
θ = cos 
 = cos 
 = 42.8°
Fs
 (30.0 N)(50.0 m) 
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−1
6.
REASONING AND SOLUTION
a. In both cases, the lift force L is perpendicular to the displacement of the plane, and, therefore,
does no work. As shown in the drawing in the text, when the plane is in the dive, there is a
component of the weight W that points in the direction of the displacement of the plane. When the
plane is climbing, there is a component of the weight that points opposite to the displacement of the
plane. Thus, since the thrust T is the same for both cases, the net force in the direction of the
displacement is greater for the case where the plane is diving. Since the displacement s is the same
in both cases, more net work is done during the dive .
b. The work done during the dive is Wdive = (T + W cos 75° ) s , while the work done during the
climb is Wclimb = (T + W cos 115° ) s . Therefore, the difference between the net work done
during the dive and the climb is
Wdive – Wclimb = ( T + W cos 75 °) s – (T + W cos 115°) s = Ws (cos 75 ° – cos 115° )
4
3
7
= (5.9 × 10 N)(1.7 × 10 m)(cos 75° – cos 115° ) = 6.8 × 10 J
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7.
REASONING The drawing shows three of the forces
+y
29.0°
that act on the cart: F is the pushing force that the
F
shopper exerts, f is the frictional force that opposes the f
+x
motion of the cart, and mg is its weight. The displacement
s of the cart is also shown. Since the cart moves at a
constant velocity along the +x direction, it is in
mg
s
equilibrium. The net force acting on it in this direction is
zero, ΣFx = 0. This relation can be used to find the magnitude of the pushing force. The work done
208 WORK AND ENERGY
by a constant force is given by Equation 6.1 as W = ( F cos θ ) s , where F is the magnitude of the
force, s is the magnitude of the displacement, and θ is the angle between the force and the
displacement.
SOLUTION
a. The x-component of the net force is zero, ΣFx = 0, so that
F cos29.0° − f = 0
144
42444
3
(4.9a)
ΣFx
The magnitude of the force that the shopper exerts is F =
f
48.0 N
=
= 54.9 N .
cos 29.0° cos 29.0°
b. The work done by the pushing force F is
W = ( F cos θ ) s = ( 54.9N )( cos29.0° )( 22.0m ) = 1060J
(6.1)
c. The angle between the frictional force and the displacement is 180°, so the work done by the
frictional force f is
W = ( f cos θ ) s = ( 48.0N ) ( cos180.0° ) ( 22.0m ) = −1060J
d. The angle between the weight of the cart and the displacement is 90°, so the work done by the
weight mg is
(
)
W = ( mg cos θ ) s = (16.0kg ) 9.80m/s 2 ( cos90 °) ( 22.0m ) = 0 J
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8.
REASONING AND SOLUTION
a. The work done by the applied force is
W = Fs cos θ = (2.40 × 102 N)(8.00 m) cos 20.0° = 1.80 × 103 J
b. The work done by the frictional force is Wf = f k s cos θ, where
f k = µ s (mg − F sin θ )
= (0.200)  (85.0 kg)(9.80 m/s2 ) − (2.40 × 102 N)(sin 20.0°)  = 1.50 × 102 N
Now
Wf = (1.50 × 102 N)(8.00 m) cos 180° = − 1.20 × 103 J
Chapter 6 Problems
209
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9.
SSM REASONING AND SOLUTION According to Equation 6.1, the work done by the
husband and wife are, respectively,
[ Husband]
[ Wife]
WH = ( FH cos θH )s
WW = ( FW cos θW )s
Since both the husband and the wife do the same amount of work,
( FH cos θH )s = ( FW cos θW )s
Since the displacement has the same magnitude s in both cases, the magnitude of the force exerted
by the wife is
cos θH
cos 58°
= (67N)
= 45N
cos θW
cos 38°
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FW = FH
10. REASONING AND SOLUTION
The work done by the force P is
WP = Ps cos 30.0° = 0.866 P s
Similarly, the work done by the frictional force is
Wf = f k s cos 180° = – f k s
where
f k = µk (mg – P sin 30.0°)
Then
Wf = – (0.200)[(1.00 × 102 kg)(9.80 m/s2) – 0.500P]s = (0.100P – 196)s
The net work is zero if
WP + Wf = 0
or
0.866 Ps + (0.100P – 196)s = 0
Solving for P gives P = 203 N .
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210 WORK AND ENERGY
11. REASONING AND SOLUTION
The net work done on the car is
WT = WF + Wf + Wg + WN
WT = Fs cos 0.0° + f s cos 180° – mgs sin 5.0° + FNs cos 90°
Rearranging this result gives
F=
WT
s
+ f + mg sin5.0°
(
)
150 × 103 J
+ 524 N + (1200kg ) 9.80m/s 2 sin5.0° = 2.07 × 103 N
290m
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=
12. REASONING AND SOLUTION
The work done on the arrow by the bow is given by
W = Fs cos 0° = Fs
This work is converted into kinetic energy according to the work energy theorem.
W = 12 mvf2 − 12 mv02
Solving for v f, we find that
vf =
2W
+ v02 =
m
2 ( 65 N ) ( 0.90 m )
–3
+ ( 0 m/s ) = 39 m/s
2
75 × 10 kg
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13.
SSM REASONING The work done to launch either object can be found from Equation 6.3,
1
2
1
2
the work-energy theorem, W = KEf − KE 0 = mvf2 − mv02 .
SOLUTION
a. The work required to launch the hammer is
(
)
W = 12 mvf2 − 12 mv02 = 12 m vf2 − v02 = 12 (7.3 kg)  (29 m/s)2 − ( 0 m/s )  = 3.1× 103 J

2


b. Similarly, the work required to launch the bullet is
Chapter 6 Problems
(
)
211
W = 12 m vf2 − v02 = 12 (0.0026 kg) (410 m/s) 2 − ( 0 m/s )  = 2.2 ×10 2 J

2


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14. REASONING The average kinetic energy is given by Equation 6.2 as K E = 12 mv 2 . The mass
of the aircraft carrier is known, and its average speed v can be obtained from Equation 2.1 as the
distance traveled divided by the elapsed time.
SOLUTION The average speed of the aircraft carrier is equal to the distance s it travels divided
by the elapsed time t, or v = s / t . Therefore, the average kinetic energy is
2
s
KE = mv = m  
t 
1
2
2
1
2
2
 208 × 10 3 m 
10
1
= 2 8.35 × 10 kg 
 = 1.14 × 10 J
 ( 3.50h)  3600s  

 1 h  
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(
15.
SSM
WWW
7
)
REASONING AND SOLUTION The work required to bring each car up
1
1
to speed is, from the work-energy theorem, W = KEf − KE 0 = mvf2 − mv02 . Therefore,
2
2
(
)
2
(
)
2
WB = 1 m vf2 − v02 = 1 (1.20 × 103 kg) (40.0 m/s)2 − ( 0 m/s )2  = 9.60 × 105 J
2


WB = 1 m vf2 − v02 = 1 (2.00 × 103 kg) (40.0 m/s)2 − ( 0 m/s ) 2  = 1.60 ×106 J
2


The additional work required to bring car B up to speed is, therefore,
6
5
5
WB − WA = (1.6 × 10 J) – (9.60 × 10 J) = 6.4 × 10 J
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16. REASONING AND SOLUTION From the work-energy theorem, Equation 6.3,
1
2
1
2
1
2
2
W = 2 mvf − 2 mv0 = 2 m vf − v0 


1
a. W = 2 (7420kg) (8450m/s) 2 –(2820m/s) 2  =


2.35 ×1011 J
212 WORK AND ENERGY
b. W = 2 (7420 kg)  (2820 m/s) 2 − (8450 m/s) 2  = –2.35 × 1011 J


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1
17. REASONING AND SOLUTION
a. The work-energy theorem gives
W = (1/2)mv f2 – (1/2)mv o2 = (1/2)(45 × 10–3 kg)(41 m/s)2 = 38 J
b. From the definition of work
W = Fs cos 0°
so
F = W/s = (38 J)/(1.0 × 10–2 m) = 3.8 × 103 N
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18. REASONING AND SOLUTION The work energy theorem, W = 12 mvf2 − 12 mv02 , gives
vf =
2W
2
+ v0
m
where
W = Fs cos 180° = − (4.0 × 105 N)(2500 × 103 m) = − 1.0 × 1012 J
Now
vf =
(
2 −1.0 × 1012 J
) + (11000m/s) 2 = 9 × 103 m/s
5.0 × 104 kg
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19.
SSM REASONING According to the work-energy theorem, the kinetic energy of the sled
increases in each case because work is done on the sled. The work-energy theorem is given by
Equation 6.3: W = KEf − KE 0 = 1 mv2f − 1 mv20 . The work done on the sled is given by
2
2
Equation 6.1: W = ( F cosθ )s . The work done in each case can, therefore, be expressed as
W1 = (F cos 0°)s =
1
2
1
2
mv2f − mv02 = ∆KE1
and
W2 = ( F cos 62 °)s = 1 mvf2 − 1 mv02 = ∆KE2
2
2
The fractional increase in the kinetic energy of the sled when θ = 0° is
Chapter 6 Problems
213
∆KE1 ( F cos 0°)s
=
= 0.38
KE0
KE0
Therefore,
Fs = (0.38) KE0
(1)
The fractional increase in the kinetic energy of the sled when θ = 62° is
∆KE2 (F cos 62°)s
Fs
=
=
(cos 62 °)
KE 0
KE0
KE0
(2)
Equation (1) can be used to substitute in Equation (2) for Fs.
SOLUTION Combining Equations (1) and (2), we have
∆KE2
Fs
(0.38) KE0
=
(cos 62°) =
(cos 62°) = (0.38)(cos 62°) = 0.18
KE 0
KE0
KE0
Thus, the sled's kinetic energy would increase by 18 % .
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20. REASONING Since the person has an upward acceleration, there must be a net force acting in
the upward direction. The net force ΣFy is related to the acceleration ay by Newton’s second law,
Σ Fy = may , where m is the mass of the person. This relation will allow us to determine the tension
in the cable. The work done by the tension and the person’s weight can be found directly from the
definition of work, Equation 6.1.
SOLUTION
a. The free-body diagram at the right shows the two forces that act on
the person. Applying Newton’s second law, we have
T
+y
s
T424
− mg
1
3 = ma y
Σ Fy
Solving for the magnitude of the tension in the cable yields
mg
T = m(ay + g) = (79 kg)(0.70 m/s2 + 9.80 m/s2) = 8.3 × 102 N
b. The work done by the tension in the cable is
WT = ( T cos θ ) s = (8.3 × 102 N) (cos 0°) (11 m) = 9.1 × 103 J
(6.1)
214 WORK AND ENERGY
c. The work done by the person’s weight is
WW = ( mg cos θ ) s = (79kg) ( 9.8m/s 2 ) (cos 180°) (11 m) = −8.5 × 103 J
(6.1)
d. The work-energy theorem relates the work done by the two forces to the change in the kinetic
energy of the person. The work done by the two forces is W = WT + WW :
WT + WW = 12 mv 2f − 12 mv 20
14
243
(6.3)
W
Solving this equation for the final speed of the person gives
vf = v02 +
(
2
W + WW
m T
)
2
( 9.1 × 103 J − 8.5 × 103 J ) = 4 m / s
79kg
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=
( 0 m/s )
2
+
21. REASONING It is useful to divide this problem into two parts. The first part involves the skier
moving on the snow. We can use the work-energy theorem to find her speed when she comes to
the edge of the cliff. In the second part she leaves the snow and falls freely toward the ground. We
can again employ the work-energy theorem to find her speed just before she lands.
SOLUTION The drawing at the right shows the three
forces that act on the skier as she glides on the snow. The
forces are: her weight mg, the normal force FN, and the
kinetic frictional force fk. Her displacement is labeled as
s. The work-energy theorem, Equation 6.3, is
FN
+y
fk
s
65.0°
25.0°
W = mv − mv
1
2
2
f
1
2
2
0
mg
where W is the work done by the net external force that
acts on the skier. The work done by each force is given
by Equation 6.1, W = ( F cos θ ) s , so the work-energy theorem becomes
( mg cos65.0o ) s + ( f k42444444444
cos180o ) s + ( FN cos90o ) s = 12 mvf2 − 12 mv02
1444444444
4
3
W
Chapter 6 Problems
215
Since cos 90° = 0, the third term on the left side can be eliminated. The magnitude f k of the kinetic
frictional force is given by Equation 4.8 as f k = µk FN . The magnitude FN of the normal force can
be determined by noting that the skier does not leave the surface of the slope, so ay = 0 m/s2.
Thus, we have that ΣFy = 0, so
FN − mg cos25.0° = 0
14442444
3
or
FN = mg cos 25.0°
Σ Fy
The magnitude of the kinetic frictional force becomes f k = µk FN = µk mg cos25.0 o . Substituting
this result into the work-energy theorem, we find that
( mg cos65.0o ) s + ( µk mg cos25.0o ) ( cos180o )3s = 12 mvf2 − 12 mv02
14444444442444444444
W
Algebraically eliminating the mass m of the skier from every term, setting cos 180° = –1 and
v 0 = 0 m/s, and solving for the final speed v f, gives
vf = 2 gs ( cos65.0o − µk cos 25.0o )
= 2 ( 9.80m/s 2 ) (10.4m ) cos65.0o − ( 0.200) cos25.0o  = 7.01m/s
The drawing at the right shows her displacement s during free fall.
Note that the displacement is a vector that starts where she leaves
the slope and ends where she touches the ground. The only force
acting on her during the free fall is her weight mg. The work-energy
theorem, Equation 6.3, is
3.50 m
θ
s
mg
W = 12 mvf2 − 12 mv02
The work W is that done by her weight, so the work-energy theorem becomes
s = 12 mv f2 − 12 mv02
( mg cos θ)3
144244
W
In this expression θ is the angle between her weight (which points vertically downward) and her
displacement. Note from the drawing that s cos θ = 3.50 m. Algebraically eliminating the mass m
of the skier from every term in the equation above and solving for the final speed v f gives
216 WORK AND ENERGY
vf = v02 + 2 g ( s cos θ )
=
( 7.01m/s ) 2 + 2 ( 9.80m/s 2 ) ( 3.50m ) = 10.9m/s
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22. REASONING AND SOLUTION Since the boat travels with constant speed, the net force on
the boat is zero. Therefore, the drive force must be equal in magnitude and opposite in direction to
the resistive force due to the water, FR. Since the force due to the water is resistive, it always
points opposite to the direction of motion of the boat and θ = 180°. After the engine is shut off,
the resistive force due to the water does negative work on the boat thereby bringing it to a halt.
a. From the work-energy theorem,
W = 12 mvf2 − 12 mv02
where W is the work done by the resistive force due to the water. Since the boat coasts to a halt,
v f = 0 m/s, and we have
( FR cos θ )s = 12 m ( 0m/s ) −
2
1
mv02
2
Solving for s gives
mv02
(480 kg)(8.9 m/s)2
s=−
=−
= 25 m
2FR cos θ
2(760 N)(cos 180°)
b. The time required for the boat to come to a halt can be found from Equation 2.4:
v = v0 + at
where v = 0 m/s since the final speed of the boat is zero. Solving for t yields
t=
( 0 m/s) − v0
a
= −
v0
F/m
= −
v0m
F
= −
(8.9 m/s)(480 kg)
= 5.6 s
( − 760 N)
Notice that the value substituted for F is negative. This is because F and v0 point in opposite
directions, and the value substituted for v 0 is positive.
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23.
SSM WWW REASONING When the satellite goes from the first to the second orbit, its
kinetic energy changes. The net work that the external force must do to change the orbit can be
1
1
found from the work-energy theorem: W = KEf − KE 0 = mvf2 − mv02 . The speeds vf and
2
2
Chapter 6 Problems
217
v0 can be obtained from Equation 5.5 for the speed of a satellite in a circular orbit of radius r.
Given the speeds, the work energy theorem can be used to obtain the work.
SOLUTION According to Equation 5.5, v = GME r . Substituting into the work-energy
theorem, we have
W =
1
2
mvf2
−
1
mv02
2
= m
1
2
(
vf2
− v02
)
 GM 2  GM
E
E
= m 
 −



rf  
r0

1
2
2




 = GM E m  1 − 1 

2  rf r0 

Therefore,
W=
(6.67×10 –11 N ⋅ m 2 /kg 2 )(5.98×10 24 kg)(6200 kg)
2
1
1
 = 1.4 × 1011 J
× 
−

6
7
 7.0 × 10 m 3.3 × 10 m 
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24. REASONING AND SOLUTION
work-energy theorem:
The net work done on the plane can be found from the
2
2
2
2
W = 12 mvf − 12 mv0 = 12 m  vf − v0 


The tension in the guideline provides the centripetal force required to keep the plane moving in a
circular path. Since the tension in the guideline becomes four times greater,
Tf = 4 T0
or
mvf2
mv02
= 4
rf
r0
Solving for vf2 gives
vf2
=
4v02 rf
r0
Substitution of this expression into the work-energy theorem gives
 4v2 r

W = m  0 f − v02  =
 r0

1
2
1 mv 2
0
2
 
 4  rf
 r
  0


 − 1




218 WORK AND ENERGY
Therefore,

1
2   14 m 
2
W = 2 (0.90 kg)(22 m/s)  4 
− 1 = 5.4 ×10 J

  16 m 

______________________________________________________________________________
25. REASONING AND SOLUTION
PE = mgh = (55.0 kg)(9.80 m/s2)(443m) = 2.39 × 105 J
(6.5)
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26. REASONING The work done by the weight of the basketball is given by
Equation 6.1 as W = ( F cos θ ) s , where F = mg is the magnitude of the weight,
s
θ is the angle between the weight and the displacement, and s is the magnitude of
the displacement. The drawing shows that the weight and displacement are parallel,
so that θ = 0°. The potential energy of the basketball is given by Equation 6.5 as mg
PE = mgh, where h is the height of the ball above the ground.
SOLUTION
a. The work done by the weight of the basketball is
W = ( F cos θ ) s = mg (cos 0°)(h0 − hf) = (0.60 kg)(9.80 m/s2)(6.1 m − 1.5 m) = 2 7 J
b. The potential energy of the ball, relative to the ground, when it is released is
PE0 = mgh0 = (0.60 kg)(9.80 m/s2)(6.1 m) = 3 6 J
(6.5)
c. The potential energy of the ball, relative to the ground, when it is caught is
PEf = mghf = (0.60 kg)(9.80 m/s2)(1.5 m) = 8.8J
(6.5)
d. The change in the ball’s gravitational potential energy is
∆PE = PEf − PE0 = 8.8 J – 36 J = −27 J
We see that the change in the gravitational potential energy is equal to –27 J = −W , where W is
the work done by the weight of the ball (see part a).
______________________________________________________________________________
27.
SSM REASONING During each portion of the trip, the work done by the resistive force is
given by Equation 6.1, W = ( F cosθ )s . Since the resistive force always points opposite to the
Chapter 6 Problems
219
displacement of the bicyclist, θ = 180°; hence, on each part of the trip,
W = ( F cos 180° )s = −Fs . The work done by the resistive force during the round trip is the
algebraic sum of the work done during each portion of the trip.
SOLUTION
a. The work done during the round trip is, therefore,
Wtotal = W1 + W2 = −F1 s1 − F2 s2
= −(3.0 N)(5.0 × 103 m) − (3.0 N)(5.0 × 103 m) = –3.0 × 10 4 J
b. Since the work done by the resistive force over the closed path is not zero, we can conclude
that the resistive force is not a conservative force .
______________________________________________________________________________
28. REASONING AND SOLUTION
a. From the definition of work, W = Fs cos θ. For upward motion θ = 180° so
W = – mgs = – (71.1 N)(2.13 m – 1.52 m) = –43 J
b. The change in potential energy is
∆PE = mghf – mgh0 = (71.1 N)(2.13 m – 1.52 m) = +43 J
______________________________________________________________________________
29. REASONING AND SOLUTION
The vertical height of the skier is h = s sin 14.6° so
s
∆PE = mgs sin 14.6°
h
14.6°
= (75.0 kg)(9.80 m/s2)(2830 m) sin 14.6°
= 5.24 × 105 J
______________________________________________________________________________
30. REASONING AND SOLUTION
The change in the pole-vaulter's potential energy is
PEf – PE0 = mghf – mgh0
Let hf = 0 m, so that h0 = 5.80 m. Then,
220 WORK AND ENERGY
PEf − PE0
0 J – 3.70 × 103 J
=
= 6.38 × 102 N
hf − h0
0 m – 5.80 m
______________________________________________________________________________
mg =
31.
SSM REASONING The only nonconservative force that acts on Rocket man is the force
generated by the propulsion unit. Thus, Equation 6.7b can be used to determine the net work
done by this force.
SOLUTION From Equation 6.7b, we have
Wnc = ∆KE + ∆PE =
(
1
2
)
mvf2 − 12 mv02 + ( mghf − mgh0 )
Since rocket man starts from rest, v 0 = 0 m/s. If we take h0 = 0 m on the ground, we have
Wnc = 1 mvf2 + mghf = m
2
(
1 v2
2 f
+ ghf
)
= (136 kg)  12 (5.0 m/s) 2 + (9.80 m/s 2 )(16 m)  = 2.3 ×10 4 J


______________________________________________________________________________
32. REASONING The only two forces that act on the gymnast are his weight and the force exerted
on his hands by the high bar. The latter is the (non-conservative) reaction force to the force
exerted on the bar by the gymnast, as predicted by Newton's third law. This force, however,
does no work because it points perpendicular to the circular path of motion. Thus, Wnc = 0 J,
and we can apply the principle of conservation of mechanical energy.
SOLUTION The conservation principle gives
1
mv f2 + mghf
214
4244
3
Ef
1
= 2 mv 02 + mgh0
144244
3
E0
Since the gymnast's speed is momentarily zero at the top of the swing, v0 = 0 m/s. If we take
hf = 0 m at the bottom of the swing, then h0 = 2r , where r is the radius of the circular path
followed by the gymnast's waist. Making these substitutions in the above expression and solving
for vf , we obtain
vf = 2gh0 = 2g(2r) = 2(9.80 m/s2 )(2 × 1.1 m) = 6.6 m/s
______________________________________________________________________________
Chapter 6 Problems
221
33. REASONING Since air resistance is being neglected, the only force that acts on the falling water
is the conservative gravitational force (its weight). Since the height of the falls and the speed of the
water at the bottom are known, we may use the conservation of mechanical energy to find the
speed of the water at the top of the falls.
SOLUTION The conservation of mechanical energy, as expressed by Equation 6.9b, states that
1 mv 2 + mgh = 1 mv 2 + mgh
f
0
2
2
144244
3f 144244
30
Ef
Ef
The mass m can be eliminated algebraically from this equation, since it appears as a factor in every
term. Solving for v 0 gives
v0 =
=
v f2 + 2 g ( hf − h0 )
( 25.8m/s )2 + 2 ( 9.80m/s 2 ) ( 0 m − 33.2m ) =
3.9m/s
______________________________________________________________________________
34. REASONING AND SOLUTION
1 mv 2
f
2
The conservation of energy gives
+ mghf =
1 mv 2
0
2
+ mgh0
Rearranging gives
hf − h0
2
2
14.0m/s ) − (13.0m/s )
(
=
= 1.4m
2 9.80m/s2
______________________________________________________________________________
35.
(
)
SSM REASONING No forces other than gravity act on the rock since air resistance is being
ignored. Thus, the net work done by nonconservative forces is zero, Wnc = 0 J. Consequently,
the principle of conservation of mechanical energy holds, so the total mechanical energy remains
constant as the rock falls.
If we take h = 0 m at ground level, the gravitational potential energy at any height h is, according to
1
Equation 6.5, PE = mgh . The kinetic energy of the rock is given by Equation 6.2: KE = mv 2 .
2
In order to use Equation 6.2, we must have a value for v 2 at each desired height h. The quantity
v 2 can be found from Equation 2.9 with v0 = 0 m/s, since the rock is released from rest. At a
height h, the rock has fallen through a distance (20.0 m) – h , and according to Equation 2.9,
v 2 = 2ay = 2a[(20.0 m) – h] . Therefore, the kinetic energy at any height h is given by
222 WORK AND ENERGY
KE = ma[ (20.0 m) – h ] . The total energy of the rock at any height h is the sum of the kinetic
energy and potential energy at the particular height.
SOLUTION The calculations are performed below for h = 10.0 m . The table that follows also
shows the results for h = 20.0 m and h = 0 m.
PE = mgh = ( 2.00 kg)(9.80 m/s 2 )(10.0 m) = 196 J
KE = ma[ (20.0 m) – h ] = (2.00 kg )(9.80 m/s 2 )(20.0 m – 10.0 m) = 196 J
E = KE + PE = 196 J + 196 J =
h (m)
20.0
10.0
0
KE (J)
0
196
392
392 J
PE (J)
392
196
0
E (J)
392
392
392
Note that in each case, the value of E is the same, because mechanical energy is conserved.
______________________________________________________________________________
36. REASONING Since air resistance is being neglected, the only force that acts on the golf ball is
the conservative gravitational force (its weight). Since the maximum height of the trajectory and the
initial speed of the ball are known, the conservation of mechanical energy can be used to find the
kinetic energy of the ball at the top of the highest point. The conservation of mechanical energy can
also be used to find the speed of the ball when it is 8.0 m below its highest point.
SOLUTION
a. The conservation of mechanical energy, Equation 6.9b, states that
1 mv 2
f
2
123
+ mghf = 12 mv02 + mgh0
KE f
Solving this equation for the final kinetic energy, KEf, yields
KE f = 12 mv02 + mg ( h0 − hf
=
b.
1
2
)
( 0.0470kg ) ( 52.0m/s ) 2 + ( 0.0470kg ) ( 9.80m/s 2 ) ( 0 m − 24.6m ) =
The conservation of mechanical energy, Equation 6.9b, states that
52.2 J
Chapter 6 Problems
1 mv 2 + mgh
f
2
144244
3f
=
223
1 mv 2 + mgh
0
2
144244
30
Ef
Ef
The mass m can be eliminated algebraically from this equation, since it appears as a factor in every
term. Solving for v f, and noting that the final height of the ball is hf = 24.6 m – 8.0 m = 16.6 m, we
have that
vf =
v 02 + 2 g ( h0 − hf
)
=
( 52.0m/s )
2
(
)
+ 2 9.80m/s 2 ( 0 m − 16.6m ) = 4 8 . 8 m / s
______________________________________________________________________________
37. REASONING AND SOLUTION
Mechanical energy is conserved if
no friction acts. Ef = E0 or
1 mv 2
f
2
+ mgh =
vf =?
h = 5.0 m
1 mv 2
0
2
Taking h0 = 0 m and rearranging
yields
vf = v02 − 2 gh =
v o = 11.0 m/s
(11.0m/s )2 − 2 ( 9.80m/s 2 ) (5.0m ) =
4.8m/s
______________________________________________________________________________
38. REASONING AND SOLUTION The hook is let go when it is h0 = 1.5 m above the ground
and traveling vertically upward. Its initial speed is
v0 =
2π r 2π (1.5 m )
=
= 28 m/s
1s
T
3
The conservation of energy requires
1 mv 2
f
2
+ mghf = 12 mv02 + mgh0
At maximum height v f = 0 m/s, so
v02
( 28 m/s ) + 1.5 m = 42 m
hf =
+ h0 =
2g
2 9.80 m/s2
2
(
)
______________________________________________________________________________
224 WORK AND ENERGY
39.
SSM WWW REASONING Gravity is the only force acting on the vaulters, since friction
and air resistance are being ignored. Therefore, the net work done by the nonconservative forces
is zero, and the principle of conservation of mechanical energy holds.
SOLUTION Let E2f represent the total mechanical energy of the second vaulter at the ground,
and E20 represent the total mechanical energy of the second vaulter at the bar. Then, the principle
of mechanical energy is written for the second vaulter as
1
mv22f + mgh2f
2144
2443
E2f
1
2
= mv20
+ mgh20
21442443
E20
Since the mass m of the vaulter appears in every term of the equation, m can be eliminated
algebraically. The quantity h20 = h , where h is the height of the bar. Furthermore, when the
vaulter is at ground level, h2 f = 0 m. Solving for v20 we have
v 20 = v 22 f − 2 gh
(1)
In order to use Equation (1), we must first determine the height h of the bar. The height h can be
determined by applying the principle of conservation of mechanical energy to the first vaulter on the
ground and at the bar. Using notation similar to that above, we have
1
2
mv1f
+ mgh1f
2144
2443
E1f
1
2
= mv10
+ mgh10
2
1442443
E10
where E10 corresponds to the total mechanical energy of the first vaulter at the bar. The height of
the bar is, therefore,
v2 − v2
(8.90 m/s)2 − (1.00 m/s) 2
h = h10 = 1f 10 =
= 3.99 m
2g
2(9.80 m/s 2 )
The speed at which the second vaulter clears the bar is, from Equation (1),
2
v20 = v 2f
− 2 gh = (9.00 m/s) 2 − 2(9.80 m/s 2 )(3.99 m) = 1.7 m/s
______________________________________________________________________________
40. REASONING AND SOLUTION Since energy is conserved, the ball will have the same
speed at the bottom of its swing whether it is moving toward or away from the crane. As the cable
swings to an angle of 32.0°, the ball will rise a distance
h = L (1 – cos 32.0°)
Chapter 6 Problems
225
where L is the length of the cable. The conservation of energy then gives
1 mv 2
0
2
= 12 mvf2 + mgh
Solving for v f gives
vf = v02 − 2 gL (1 − cos32.0° )
=
( 6.50m/s )2 − 2 ( 9.80m/s2 ) (10.0m )( 1 − cos32.0°) =
3.53m/s
______________________________________________________________________________
41.
SSM REASONING Friction and air resistance are being ignored. The normal force from the
slide is perpendicular to the motion, so it does no work. Thus, no net work is done by
nonconservative forces, and the principle of conservation of mechanical energy applies.
SOLUTION Applying the principle of conservation of mechanical energy to the swimmer at the
top and the bottom of the slide, we have
1
mv2f + mghf
2 4
1
4244
3
Ef
1
= mv20 + mgh0
2 4
1
4244
3
E0
If we let h be the height of the bottom of the slide above the water, hf = h , and h0 = H . Since
the swimmer starts from rest, v0 = 0 m/s, and the above expression becomes
1 2
v
2 f
+ gh = gH
Solving for H, we obtain
v2
H = h+ f
2g
Before we can calculate H, we must find vf and h. Since the velocity in the horizontal direction is
constant,
∆x 5.00 m
vf =
=
= 10.0 m/s
∆t 0.500 s
The vertical displacement of the swimmer after leaving the slide is, from Equation 3.5b (with down
being negative),
y = 2 ay t 2 = 2 (–9.80 m/s 2 )( 0.500 s) 2 = −1.23 m
1
1
Therefore, h = 1.23 m. Using these values of vf and h in the above expression for H, we find
226 WORK AND ENERGY
v2
(10.0 m/s) 2
H = h + f = 1.23 m +
= 6.33 m
2g
2(9.80 m/s 2 )
______________________________________________________________________________
42. REASONING AND SOLUTION If air resistance is ignored, the only nonconservative force
that acts on the skier is the normal force exerted on the skier by the snow. Since this force is
always perpendicular to the direction of the displacement, the work done by the normal force is
zero. We can conclude, therefore, that mechanical energy is conserved.
1
mv 20 +
2
1
mgh 0 = mv 2f + mgh f
2
Since the skier starts from rest v0 = 0 m/s. Let hf define the zero level for heights, then the final
gravitational potential energy is zero. This gives
mgh 0 =
1
2
mv 2f
(1)
At the crest of the second hill, the two forces that act on the skier are
the normal force and the weight of the skier. The resultant of these two
forces provides the necessary centripetal force to keep the skier moving
along the circular arc of the hill. When the skier just loses contact with
the snow, the normal force is zero and the weight of the skier must
provide the necessary centripetal force.
mg =
mv 2f
r
v 2f =gr
so that
F
N
mg
(2)
Substituting this expression for v 2f into Equation (1) gives
mgh 0 =
1
2
mgr
Solving for h0 gives
r 36 m
=
= 18 m
2
2
______________________________________________________________________________
h0 =
43. REASONING AND SOLUTION
At the bottom of the circular path of the swing, the
centripetal force is provided by the tension in the rope less the weight of the swing and rider. That
is,
mv 2
= T − mg
r
{
FC
Chapter 6 Problems
227
Solving for the mass yields
m=
T
v2
+g
r
The energy of the swing is conserved if friction is ignored. The initial energy, E0, when the swing is
released is completely potential energy and is E0 = mgh0, where
h0 = r (1 – cos 60.0°) = 12 r
is the vertical height of the swing. At the bottom of the path the swing's energy is entirely kinetic
and is
Ef = 12 mv2
Setting Eo = Ef and solving for v gives
v = 2gh0 = gr
The expression for the mass now becomes
8.00 × 102 N
T
=
= 40.8kg
2 g 2 9.80m/s 2
______________________________________________________________________________
m=
(
)
44. REASONING AND SOLUTION If air resistance is ignored, the only nonconservative force
that acts on the person is the normal force exerted on the person by the surface. Since this force is
always perpendicular to the direction of the displacement, the work done by the normal force is
zero. We can conclude, therefore, that mechanical energy is conserved.
1
mv 20 +
2
1
2
mgh 0 = mv 2f + mgh f
(1)
228 WORK AND ENERGY
where the final state pertains to the position where the
person leaves the surface. Since the person starts from
rest v0 = 0 m/s. Since the radius of the surface is r,
h0 = r, and hf = r cos θf where θf is the angle at which the
person leaves the surface.
r
θf
r cos θf
r
Equation (1) becomes
1
mgr = mv 2f + mg( r cos θf )
2
(2)
In general, as the person slides down the surface, the two
forces that act on him are the normal force FN and the weight
mg.
The centripetal force required to keep the person
moving in the circular path is the resultant of FN and the radial
component of the weight, mg cos θ.
When the person leaves the surface, the normal force is zero,
and the radial component of the weight provides the
centripetal force.
mg cos θf =
mv 2f
r
F
N
mg cos θ
θ
θ
θ
mg
v2f = gr cos θf (3)
⇒
Substituting this expression for v 2f into Equation (2) gives
mgr = 12 mg (r cos θf ) + mg (r cos θf )
Solving for θf gives
2
θf = cos −1   = 48°
3
______________________________________________________________________________
45.
SSM
REASONING AND SOLUTION The work-energy theorem can be used to
determine the change in kinetic energy of the car according to Equation 6.8:
Wnc =
( mv + mgh ) − ( mv + mgh )
1442443 1442443
1
2
2
f
f
Ef
1
2
2
0
0
E0
The nonconservative forces are the forces of friction and the force due to the chain mechanism.
Thus, we can rewrite Equation 6.8 as
Chapter 6 Problems
229
Wfriction + Wchain = ( KEf + mghf ) − ( KE 0 + mgh0 )
We will measure the heights from ground level. Solving for the change in kinetic energy of the car,
we have
KEf − KE0 = Wfriction + Wchain − mghf + mgh0
= − 2.00×104 J + 3.00×10 4 J − (375 kg)(9.80 m/s 2 )(20.0 m)
+ (375 kg)(9.80 m/s 2 )(5.00 m)= –4.51×104 J
______________________________________________________________________________
46. REASONING AND SOLUTION
The work done by air resistance is equal to the
energy lost by the ball.
2.00 m
Wnc = Ef – E0
Wnc =
Wnc =
(
1
2
) (
2
mvf + mghf −
1 (0.600
2
1
2
2
mv0 + mgh0
3.10 m
)
kg)(4.20 m/s)2 + (0.600 kg)(9.80 m/s2)(3.10 m)
– 12 (0.600 kg)(7.20 m/s)2 – (0.600 kg)(9.80 m/s2)(2.00 m) = −3.8J
______________________________________________________________________________
47. REASONING AND SOLUTION
a. The work done by non-conservative forces is given by Equation 6.7b as
Wnc = ∆KE + ∆PE
∆PE = Wnc – ∆KE
so
Now
∆KE =
1
2
mvf2 −
1
2
mv02 =
1
2
(55.0 kg)[(6.00 m/s)2 − (1.80 m/s) 2 ] = 901 J
and
∆PE = 80.0 J – 265 J – 901 J = −1086 J
b.
∆PE = mg (h – h0) so
h – h0 =
∆ PE
–1086 J
=
= –2.01 m
mg
(55.0 kg ) 9.80 m/s 2
(
)
230 WORK AND ENERGY
Thus, the skater’s vertical position has changed by 2.01 m , and the skater is
below the starting point .
______________________________________________________________________________
48. REASONING
a. Since there is no air friction, the only force that acts on the projectile is the conservative
gravitational force (its weight). The initial and final speeds of the ball are known, so the
conservation of mechanical energy can be used to find the maximum height that the projectile
attains.
b. When air resistance, a nonconservative force, is present, it does negative work on the projectile
and slows it down. Consequently, the projectile does not rise as high as when there is no air
resistance. The work-energy theorem, in the form of Equation 6.6, may be used to find the work
done by air friction. Then, using the definition of work, Equation 6.1, the average force due to air
resistance can be found.
SOLUTION
a. The conservation of mechanical energy, as expressed by Equation 6.9b, states that
1 mv2 + mgh = 1 mv 2 + mgh
f
0
2
2
144244
3f 144244
30
Ef
E0
The mass m can be eliminated algebraically from this equation since it appears as a factor in every
term. Solving for the final height hf gives
hf =
1
2
( v02 − vf2 ) + h
0
g
Setting h0 = 0 m and v f = 0 m/s, the final height, in the absence of air resistance, is
hf =
vo2 − vf2
2g
=
(18.0m/s ) 2 − ( 0 m/s ) 2
(
2 9.80m/s2
)
= 16.5m
b. The work-energy theorem is
Wnc =
(
1 mv 2
f
2
2
− 12 mv0
) + ( mgh
f
− mgh0 )
(6.6)
Chapter 6 Problems
231
where Wnc is the nonconservative work done by air resistance. According to Equation 6.1, the
work can be written as Wnc = ( FR cos180° ) s , where FR is the average force of air resistance.
As the projectile moves upward, the force of air resistance is directed downward, so the angle
between the two vectors is θ = 180° and cos θ = –1. The magnitude s of the displacement is the
difference between the final and initial heights, s = hf – h0 = 11.8 m. With these substitutions, the
work-energy theorem becomes
(
)
− FR s = 12 m vf2 − vo2 + mg ( hf − h0 )
Solving for FR gives
FR =
=
1
2
(
)
m vf2 − vo2 + mg ( hf − h0 )
−s
1
2
( 0.750kg ) ( 0m/s ) 2 − (18.0m/s ) 2  + ( 0.750kg ) ( 9.80m/s 2 ) (11.8m)
=
− (11.8m)
2.9N
______________________________________________________________________________
49. REASONING AND SOLUTION The force exerted by the bat on the ball is the only nonconservative force acting. The work due to this force is
Wnc = mv f2 – mv02 + mg ( hf – h0 )
1
1
2
2
Taking h0 = 0 m at the level of the bat, v 0 = 40.0 m/s just before the bat strikes the ball and v f to
be the speed of the ball at hf = 25.0 m, we have
vf =
2Wnc
m
+ v02 – 2ghf
2(70.0 J)
2
+ ( 40.0 m/s ) –2(9.80 m/s2 )(25.0 m) = 45.9 m/s
0.140 kg
______________________________________________________________________________
=
50. REASONING
a. Because the kinetic frictional force, a nonconservative force, is present, it does negative work
on the skier. The work-energy theorem, in the form of Equation 6.6, may be used to find the work
done by this force.
232 WORK AND ENERGY
b. Once the work done by the kinetic frictional force is known, the magnitude of the kinetic
frictional force can be determined by using the definition of work, Equation 6.1, since the
magnitude of the skier’s displacement is known.
SOLUTION
a. The work Wnc done by the kinetic frictional force is related to the object’s kinetic and potential
energies by Equation 6.6:
Wnc =
(
1
2
mv2f − 12 mv20
) + ( mgh
− mgh0 )
f
The initial height of the skier at the bottom of the hill is h0 = 0 m, and the final
height is hf = s sin 25° (see the drawing). Thus, the work is
Wnc =
(
=
1
2
1 mv 2
f
2
− 12 mv02
s
hf
25°
) + mg ( s sin 25° − h )
0
( 63kg ) ( 4 . 4 m / s ) 2 − 12 ( 63kg )( 6 . 6 m / s ) 2
(
)
+ ( 63kg ) 9.80m/s 2 (1.9m ) sin25° − 0 m  = −2 7 0 J
b. The work done by the kinetic frictional force is, according to Equation 6.1,
Wnc = ( f k cos180° ) s , where f k is the magnitude of the kinetic frictional force, and s is the
magnitude of the skier’s displacement. The displacement of the skier is up the hill and the kinetic
frictional force is directed down the hill, so the angle between the two vectors is θ = 180° and cos
θ = –1. Solving the equation above for f k, we have
Wnc −270J
=
= 140N
−s −1.9m
______________________________________________________________________________
fk =
51. REASONING The work-energy theorem, as expressed in Equation 6.7b, is
W nc = ∆ KE + ∆ PE
and can be applied to method 1 and method 2. In either method, the speed is constant, so that
∆KE = 0 J. In addition, we take the height at ground level to be h0 = 0 m, so that hf = h and
∆ PE = mghf – mgh0 = mgh . Thus, the work-energy theorem becomes
Wnc = mgh
(1)
Chapter 6 Problems
233
SOLUTION In method 1, the nonconservative work that you do is the only nonconservative
work present, so that Equation (1) becomes
Wnc = W you, 1 = mgh
In method 2, there are two contributions to the nonconservative work, so that Equation (1)
becomes
Wnc = –485 J + Wyou, 2 = mgh
Since Wyou, 2 = 2Wyou, 1 = 2mgh, we have
–485 J + 2 mgh = mgh
485 J = mgh
or
But h = (3.00 m) sin θ, so it follows that
485 J = mg(3.00 m ) sin θ
Therefore,


 485 J

485 J
–1 
 = 21.5°
θ = sin 
 = sin 
2

 mg ( 3.00 m ) 
 ( 45.0kg ) 9.80 m/s ( 3.00 m ) 
______________________________________________________________________________
–1
(
)
52. REASONING AND SOLUTION
a. The lost energy is
Elost = E0 – Ef
The ball dropped from rest, so its initial energy is purely potential. The ball is momentarily at rest
at the highest point in its rebound, so its final energy is also purely potential. Then
(
)
2
Elost = mgh0 – mghf = ( 0.60kg ) 9.80 m/s  (1.22 m ) – ( 0.69 m )  = 3.1 J
b. The work done by the player must compensate for this loss of energy.
Elost
3.1 J
= 24 N
(cos θ) s
(cos 0°)(0.13 m)
______________________________________________________________________________
Elost = (F cos θ) s
53.
⇒
F=
=
SSM REASONING AND SOLUTION According to the work-energy theorem as given in
Equation 6.8, we have
234 WORK AND ENERGY
Wnc =
(
1
mvf2
2
+ mghf
)−(
1
2
mv02 + mgh0
)
The metal piece starts at rest and is at rest just as it barely strikes the bell, so that vf = v0 = 0 m/s.
(
2
)
In addition, hf = h and h0 = 0 m, while Wnc = 0.25 12 M v , where M and v are the mass and
speed of the hammer. Thus, the work-energy theorem becomes
0.25
(
1 M v2
2
) = mgh
Solving for the speed of the hammer, we find
2mgh
2(0.400 kg)(9.80 m/s 2 )(5.00 m)
=
= 4.17 m/s
0.25M
0.25 (9.00 kg)
______________________________________________________________________________
v =
54. REASONING AND SOLUTION
For the actual motion of the rocket
∆KE = Wnc – ∆PE = Wnc – mgh
(6.7b)
∆KE = –8.00 × 102 J – (3.00 kg)(9.80 m/s2)(1.00 × 102 m) = –3740 J
Since ∆KE = KEf – KE0 = – KE0 , the initial kinetic energy of the rocket is then
KE0 = 3740 J
If the rocket were launched with this initial kinetic energy and no air resistance, the potential energy
of the rocket at the top of its trajectory would be
PE = mgh = 3740 J
Hence,
h=
3740J
( 3.00kg ) ( 9.80m/s2 )
= 127m
______________________________________________________________________________
55. REASONING
We can find the power produced by the hero by using Equation 6.10a,
P = W / t . In order to use this expression, we must first determine the work done by the hero. In
order to calculate the work, we must first find the net force. The net force can be found from
Newton's second law.
SOLUTION There are two forces that act on the villain while he is being lifted upward. They
are the upward vertical force F exerted on the villain by the hero, and the downward weight mg.
Chapter 6 Problems
235
With upward taken as positive, Newton’s second law gives F − mg = ma = 0 , since the villain
moves straight upward at constant speed. Therefore, F = mg , and the work done by the hero in
lifting the villain straight up through a vertical distance h is W = Fh = mgh . Therefore, the power
produced by the hero is
W mgh (91 kg)(9.80 m/s 2 )(1.2 m)
=
=
= 2.1 × 103 W
t
t
0.51 s
______________________________________________________________________________
P=
56. REASONING AND SOLUTION
The work done by the person in turning the crank one
revolution is
W = F (2πr) = (22 N)(2π)(0.28 m) = 39 J
The power expended is then
W 39J
=
= 3.0 × 101 W
(6.10a)
t 1.3s
______________________________________________________________________________
P=
57.
SSM REASONING AND SOLUTION One kilowatt ⋅ hour is the amount of work or
energy generated when one kilowatt of power is supplied for a time of one hour. From
Equation 6.10a, we know that W = P t . Using the fact that 1 kW = 1.0 × 10 3 J/s and that
1h = 3600 s, we have
3
3
6
1.0 kWh = (1.0 × 10 J/s)(1 h) = (1.0 × 10 J/s)(3600 s) = 3.6 × 10 J
______________________________________________________________________________
58. REASONING AND SOLUTION
gravity, W = mgh, so
The work done by the crane is the work done against
(
)(
)
3.00 × 102 kg 9.80m/s2 (10.0m)
mgh
t=
=
= 73.5s
P
4.00 × 102 W
______________________________________________________________________________
59. REASONING The average power developed by the cheetah is equal to the work done by the
cheetah divided by the elapsed time (Equation 6.10a). The work, on the other hand, can be related
to the change in the cheetah’s kinetic energy by the work-energy theorem, Equation 6.3.
SOLUTION
a. The average power is
P=
W
t
(6.10a)
236 WORK AND ENERGY
where W is the work done by the cheetah. This work is related to the change in the cheetah’s
kinetic energy by Equation 6.3, W = 12 mvf2 − 12 mv02 , so the average power is
W
P=
=
t
=
mvf2 − 12 mv02
t
1 (110kg )( 2 7 m / s )2 −
2
1
2
1
2
( 110kg ) ( 0 m / s )2
4.0s
= 1.0 × 104 W
b. The power, in units of horsepower (hp), is
(
)
 1hp 
4
P = 1.0 × 10 W 
 = 13hp
 745.7W 
______________________________________________________________________________
60. REASONING AND SOLUTION
a. The power developed by the engine is
P = Fv = (2.00 × 102 N)(20.0 m/s) = 4.00 × 103 W
b. The force required of the engine in order to maintain a constant speed up the slope is
F = Fa + mg sin 37.0°
The power developed by the engine is then
P = Fv = (Fa + mg sin 37.0°)v
(6.11)
P = [2.00 × 102 N + (2.50 × 102 kg)(9.80 m/s2)sin 37.0°](20.0 m/s) = 3.35 × 10 4 W
______________________________________________________________________________
61.
SSM REASONING AND SOLUTION In the drawings below, the positive direction is to
the right. When the boat is not pulling a skier, the engine must provide a force F1 to overcome
the resistive force of the water FR. Since the boat travels with constant speed, these forces must
be equal in magnitude and opposite in direction.
FR
– FR + F1 = ma = 0
or
FR = F1
(1)
F1
Chapter 6 Problems
237
When the boat is pulling a skier, the engine must provide a force F2 to balance the
resistive force of the water FR and the tension in the tow rope T.
T
F
– FR + F2 – T = ma = 0
R
F
2
or
F2 = FR + T
(2)
The average power is given by P = Fv , according to Equation 6.11 in the text. Since the boat
moves with the same speed in both cases, v 1 = v 2, and we have
P1 P 2
=
F1 F2
or
F2 = F1
P2
P1
Using Equations (1) and (2), this becomes
F1 + T = F1
P2
P1
Solving for T gives
P

T = F1  2 − 1
 P1

The force F1 can be determined from P1 = F1v1 , thereby giving
 7.50 × 104 W  8.30 × 104 W 
P1  P 2
−
1
− 1  = 6.7 × 102 N


=
4
v1  P1
12m/s  7.50 × 10 W 

______________________________________________________________________________
T=
62. REASONING AND SOLUTION The following drawings show the free-body diagrams for
the car in going both up and down the hill. The force FR is the combined force of air resistance
and friction, and the forces FU and FD are the forces supplied by the engine in going uphill and
downhill respectively.
238 WORK AND ENERGY
Going up the hill
FN
Going down the hill
FN
FU
FR
FD
FR
mg sin θ
mg cos θ
mg sin θ
mg cos θ
Writing Newton's second law in the direction of motion for the car as it goes uphill, taking uphill as
the positive direction, we have
FU − FR − mg sin θ = ma = 0
Solving for FU , we have
FU = FR + mg sin θ
Similarly, when the car is going downhill, Newton's second law in the direction of motion gives
FR − FD − mg sin θ = ma = 0
so that
FD = FR − mg sin θ
Since the car needs 47 hp more to sustain the constant uphill velocity than the constant downhill
velocity, we can write
P U = P D + ∆P
where PU is the power needed to sustain the constant uphill velocity, PD is the power needed to
sustain the constant downhill velocity, and ∆ P = 47 hp . In terms of SI units,
 746 W 
∆P = ( 47 hp ) 
 = 3.51 ×10 4 W
 1 hp 
Using Equation 6.11 ( P = Fv ), the equation PU = P D + ∆P can be written as
FU v = FD v + ∆P
Using the expressions for FU and FD , we have
Solving for θ, we find
( FR + mg sin θ ) v = ( FR − mg sin θ ) v + ∆ P
Chapter 6 Problems
239

∆P 
3.51 ×10 4 W
–1 
θ = sin 
 = 2.0°
 = sin 
2
 2 mgv 
 2(1900 kg)(9.80 m/s )(27 m/s) 
______________________________________________________________________________
–1 
63.
SSM REASONING The area under the force-versus-displacement graph over any
displacement interval is equal to the work done over that displacement interval.
SOLUTION
a. Since the area under the force-versus-displacement graph from s = 0 to 0.50 m is greater for
bow 1, it follows that bow 1 requires more work to draw the bow fully
.
b. The work required to draw bow 1 is equal to the area of the triangular region under the forceversus-displacement curve for bow 1. Since the area of a triangle is equal one-half times the base
of the triangle times the height of the triangle, we have
W1 = 1 (0.50m)(350 N) = 88 J
2
For bow 2, we note that each small square under the force-versus-displacement graph has an area
of
(0.050 m)(40.0 N) = 2.0 J
We estimate that there are approximately 31.3 squares under the force-versus-displacement graph
for bow 2; therefore, the total work done is
 2.0 J 
(31.3 squares) 
= 63 J
 square 
Therefore, the additional work required to stretch bow 1 as compared to bow 2 is
88 J – 63 J =
25 J
______________________________________________________________________________
64. REASONING AND SOLUTION The work done during each interval is equal to the area
under the force vs. displacement curve over that interval.
a. Since the area under the curve from 0 to 1.0 m is triangular, the area under the curve over this
interval is (1/2) × height × base:
W01 = 1 ( +6.0N)(1.0m − 0 m ) = 3.0 J
2
240 WORK AND ENERGY
b. Since there is no area under the curve, the
work done is zero . This is reasonable since
there is no net force acting on the object during this interval.
c. In the interval from 2.0 to 4.0 m, the area under the curve has two distinct regions. From 2.0
to 3.0 m, the area is triangular in shape, while from 3.0 to 4.0 m the region is rectangular. The total
work done from 2.0 to 4.0 m is the sum of the areas of these two regions.
1
W 23 = 2 (–6.0 N)(3.0 m – 2.0 m) = – 3.0 J
and
W 34 = (–6.0 N)(4.0 m – 3.0 m) = – 6.0 J
Thus, the total work done in the interval from 2.0 to 4.0 m is
W24 = W23 + W34 = (–3.0 J) + (–6.0 J) =
–9.0 J
Notice in the interval from 2.0 to 4.0 m, the area under the curve (and hence the work done) is
negative since the force is negative in that interval.
______________________________________________________________________________
65. REASONING AND SOLUTION
a. The work done is equal to the colored area under the graph. Therefore, the work done by the
net force on the skater from 0 to 3.0 m is
 Area under graph  = ( 31 N − 0 N ) ( 3.0 m − 0 m ) = 93 J
 from 0 to 3.0 m 
b. From 3.0 m to 6.0 m, the net force component F cos θ is zero, and there is no area under the
curve. Therefore, no work is done .
c. The speed of the skater will increase from 1.5 m/s when s = 0 m to a final value vf at
s = 3.0 m . Since the force drops to zero at s = 3.0 m , the speed remains constant at the value
vf from s = 3.0 m to s = 6.0 m ; therefore, the speed at s = 6.0 m is vf . We can find vf from
the work-energy theorem (Equation 6.3). According to the work-energy theorem, the work done
on the skater is given by
1
2
1
2
W = mvf2 − mv02
Solving for vf , we find
2W
2(93 J)
+ v02 =
+ (1.5 m/s) 2 = 2.3 m/s
m
65 kg
______________________________________________________________________________
vf =
Chapter 6 Problems
241
66. REASONING The area under the force-versus-displacement graph over any displacement
interval is equal to the work done over that interval. From Example 16, we know that the total
work done in drawing back the string of the bow is 60.5 J, corresponding to a total area under the
curve of 242 squares (each square represents 0.250 J of work). The percentage of the total work
done over any displacement interval is, therefore,
Work done during interval
Number of squares under curve during interval
×100 =
× 100
Total work done during all intervals
Total number of squares under entire curve
SOLUTION
a. We estimate that there are 130 squares under the curve from s = 0 to 0.306 m ; therefore, the
percentage of work done during the interval in question is
130 squares
×100 = 54%
242 squares
b. Similarly, we estimate that there are 112 squares under the curve in the interval from
s = 0.306 m to 0.500 m ; therefore, the percentage of work done during the interval in question is
112 squares
× 100 = 46%
242 squares
______________________________________________________________________________
67. REASONING AND SOLUTION
a. The net work done is equal to the area under the curve. For the first 10 m, the area is triangular
and the area under the curve is equal to (1/2) × height × base; therefore
W0,10 = 12 (10.0N)(10.0m − 0 m )= 5.00× 101 J
From 10 to 20 m the region is rectangular and the area under the curve is
W10,20 = (10.0N)(20.0m − 10.0m) = 1.00 ×102 J
Thus, the total work done by the force is
W0,10 + W10,20 = (5.00 × 101 J) + (1.00 × 102 J) = 1.50 × 102 J
b. From the work-energy theorem,
W = 12 mvf2 − 12 mv02
Since the object is initially at rest, v 0 = 0 m/s. Solving for v f gives
242 WORK AND ENERGY
2W
2(1.50 ×10 2 J)
vf =
=
= 7.07 m/s
m
6.00 kg
______________________________________________________________________________
68. REASONING The work done by the tension in the cable is given by Equation 6.1 as
W = ( T cos θ ) s . Since the elevator is moving upward at a constant velocity, it is in equilibrium,
and the magnitude T of the tension must be equal to the magnitude mg of the elevator’s weight; T
= mg.
SOLUTION
a. The tension and the displacement vectors point in the same direction (upward), so the angle
between them is θ = 0°. The work done by the tension is
W = ( T cos θ ) s = ( mg cos θ ) s
(6.1)
= (1200kg ) (9.80m/s 2 ) cos0° ( 3 5 m ) = 4.1 × 10 5 J
b. The weight and the displacement vectors point in opposite directions, so the angle between
them is θ = 180°. The work done by the weight is
W = ( mg cos θ ) s = (1200kg ) ( 9.80m/s 2 ) cos180° ( 3 5 m ) = −4.1 × 10 5 J
(6.1)
______________________________________________________________________________
69.
SSM REASONING AND SOLUTION Since we are ignoring friction and air resistance,
the net work done by the nonconservative forces is zero, and the principle of conservation of
mechanical energy holds. Let E0 be the total mechanical energy of the vaulter at take-off and Ef
be the total mechanical energy as he clears the bar. The principle of conservation of mechanical
energy gives
1
mv2f + mghf = 1 mv20 + mgh0
214
2 4
4244
3 1
4244
3
Ef
E0
For maximum height hf , the vaulter just clears the bar with zero speed, vf = 0 m/s. If we let
h0 = 0 m at ground level, we have
1
2
mghf = mv20
Solving for hf , we find
v02
(9.00 m/s) 2
hf =
=
= 4.13 m
2 g 2(9.80 m/s 2 )
______________________________________________________________________________
Chapter 6 Problems
70. REASONING AND SOLUTION
243
The work done by the pitcher is
Wnc = F (π r)
Assuming that only the pitcher does work on the ball
Wnc =
1
2
mvf2 − 12 mv02 + mghf − mgh0
Then,
vf =
=
2Wnc
m
+ v02 – 2 g ( hf – h0 )
2(28 N) π ( 0.51 m )
+ (12 m/s ) – 2(9.80 m/s2 )(–1.02 m) = 23 m/s
2
0.25 kg
______________________________________________________________________________
71.
SSM REASONING AND SOLUTION The work done by the retarding force is given by
Equation 6.1:W = ( F cosθ )s . Since the force is a retarding force, it must point opposite to the
direction of the displacement, so that θ = 180°. Thus, we have
3
6
W = ( F cosθ )s = (3.0 × 10 N)(cos 180 °)(850 m) = –2.6 × 10 J
The work done by this force is negative
, because the retarding force is directed opposite to
the direction of the displacement of the truck.
______________________________________________________________________________
72. REASONING AND SOLUTION The work done by nonconservative forces on the balloon is
Wnc = ∆KE + ∆PE = +9.70 × 104 J
The change in kinetic energy of the balloon is
∆KE =
1 mv 2
f
2
=
1 (5.00
2
× 102 kg)(8.00 m/s)2 = +1.60 × 104 J
The change in the potential energy is
∆PE = mgh = (5.00 × 102 kg)(9.80 m/s2)h = (4.90 × 103 kg⋅m/s2)h
Hence,
h=
9.70 × 104 J − 1.60 × 104 J
= 16.5m
4.90 × 103 kg ⋅ m/s 2
______________________________________________________________________________
244 WORK AND ENERGY
73. REASONING AND SOLUTION
The applied force does work
WP = Ps cos 0° = (150 N)(7.0 m) = 1.0 × 103 J
The frictional force does work
Wf = f ks cos 180° = – µkFNs
where FN = mg, so
Wf = – (0.25)(55 kg)(9.80 m/s2)(7.0 m) = –940 J
The normal force and gravity do no work , since they both act at a 90° angle to the
displacement.
______________________________________________________________________________
74. REASONING The average power generated by the tension in the cable is equal to the work
done by the tension divided by the elapsed time (Equation 6.10a). The work can be related to the
change in the lift’s kinetic and potential energies via the work-energy theorem.
SOLUTION
a. The average power is
P=
Wnc
t
(6.10a)
where Wnc is the work done by the nonconservative tension force. This work is related to the lift’s
kinetic and potential energies by Equation 6.6, Wnc =
(
1
2
)
mvf2 − 12 mv02 + ( mghf − mgh0 ) , so the
average power is
W
P = nc =
t
(
1
2
mvf2 − 12 mv02
) + (mgh
f
− mgh0 )
t
Since the lift moves at a constant speed, v 0 = v f, the average power becomes
( 4skiers )  65
(
)
kg 
9.80m/s 2 (1 4 0 m)

mg ( hf − h0 )
 skier 
P=
=
= 3.0 × 103 W
t
 60s 
( 2 min) 

 1min 
______________________________________________________________________________
Chapter 6 Problems
75.
245
SSM REASONING The work-energy theorem can be used to determine the net work done
on the car by the nonconservative forces of friction and air resistance. The work-energy theorem
is, according to Equation 6.8,
Wnc =
(
1 mv 2
f
2
) (
+ mghf −
1
2
2
mv0 + mgh0
)
The nonconservative forces are the force of friction, the force of air resistance, and the force
provided by the engine. Thus, we can rewrite Equation 6.8 as
Wfriction + Wair + Wengine =
(
1
2
) (
2
mvf + mghf −
2
mv0 + mgh0
1
2
)
This expression can be solved for Wfriction + Wair .
SOLUTION We will measure the heights from sea level, where h0 = 0 m. Since the car starts
from rest, v0 = 0 m/s. Thus, we have
Wfriction + Wair = m
(
1 v2
2 f
)
+ ghf − Wengine
= (1.50 × 103 kg)  12 (27.0 m/s)2 + (9.80 m/s2 )(2.00× 102 m) – (4.70 × 10 6 J)


= –1.21 ×10 6 J
______________________________________________________________________________
76. REASONING AND SOLUTION Relative to her final position in the water, the diver initially
had energy mgh0. All of this energy was dissipated as work in traveling through 3.00 m of water.
Wnc =
1 mv 2
f
2
–
1 mv 2
0
2
+ mghf – mgh0
Thus,
Wnc = Fs cos 180° = mghf – mgh0
Then,
( 55.0kg ) ( 9.80 m/s 2 ) ( –13.0 m )
F=
=
( 3.00 m ) cos180°
2340 N
______________________________________________________________________________
77. REASONING AND SOLUTION The conservation of energy applied between point A and
the top of the trajectory gives
246 WORK AND ENERGY
KEA + mghA = mgh
where h = 4.00 m. Rearranging, we find
KEA = mg(h – hA )
or
(
)
vA = 2g ( h − hA ) = 2 9.80m/s 2 ( 4.00m − 3.00m ) = 4.43m/s
______________________________________________________________________________
78. REASONING AND SOLUTION
consists of the full weight of the car.
When the car is at the top of the track the centripetal force
mv2/r = mg
Applying the conservation of energy between the bottom and the top of the track gives
(1/2)mv2 + mg(2r) = (1/2)mv02
Using both of the above equations
v02 = 5gr
so
r = v02/(5g) = (4.00 m/s)2/(49.0 m/s2) = 0.327 m
______________________________________________________________________________
79.
SSM WWW REASONING After the wheels lock, the only nonconservative force acting
on the truck is friction. The work done by this conservative force can be determined from the
work-energy theorem. According to Equation 6.8,
Wnc = E f − E0 =
(
1
2
2
) (
mvf + mghf −
1
2
2
mv0 + mgh0
)
(1)
where Wnc is the work done by friction. According to Equation 6.1, W = ( F cosθ )s ; since the
force of kinetic friction points opposite to the displacement, θ = 180°. According to Equation 4.8,
the kinetic frictional force has a magnitude of f k = µk FN , where µk is the coefficient of kinetic
friction and FN is the magnitude of the normal force. Thus,
Wnc = ( f k cosθ ) s = µk FN (cos 180°) s = −µk FN s
(2)
Chapter 6 Problems
Since the truck is sliding down an incline, we refer to the freebody diagram in order to determine the magnitude of the normal
force FN . The free-body diagram at the right shows the three
forces that act on the truck. They are the normal force FN , the
force of kinetic friction f k , and the weight mg . The weight has
been resolved into its components, and these vectors are shown
as dashed arrows. From the free body diagram and the fact that
the truck does not accelerate perpendicular to the incline, we can
see that the magnitude of the normal force is given by
247
fk
FN
mg 15°
mg cos 15°
mg sin 15°
FN = mg cos 15°
Therefore, Equation (2) becomes
Wnc = − µk (mg cos 15°) s
(3)
Combining Equations (1), (2), and (3), we have
− µk ( mg cos 15° ) s =
(
1
2
2
mvf + mghf
)−(
1
2
2
mv0 + mgh0
)
(4)
This expression may be solved for s.
SOLUTION When the car comes to a stop, vf = 0 m/s. If we
take hf = 0 m at ground level, then, from the drawing at the right,
we see that h0 = s sin 15° . Equation (4) then gives
s
h0
15°
v20
(11.1 m/s) 2
=
= 13.5 m
2g(µk cos 15° – sin 15 ° ) 2(9.80 m/s 2 )[(0.750)(cos 15 ° ) – sin 15 °]
______________________________________________________________________________
s=
80. CONCEPT QUESTIONS
a. Since the force exerted by the elevator and the displacement are in the same direction on the
upward part of the trip, the angle between them is 0°, and so the work done by the force is
positive.
b. Since the force exerted by the elevator and the displacement are in opposite directions on the
downward part of the trip, the angle between them is 180°, and so the work done by the force is
negative.
SOLUTION
FN
+y
W
W
248 WORK AND ENERGY
a.
The free-body diagram at the right shows the three forces that act on you: W is your
weight, Wbelongings is the weight of your belongings, and FN is the normal force exerted on you
by the floor of the elevator. Since you are moving upward at a constant velocity, you are in
equilibrium and the net force in the y direction must be zero:
FN − W − Wbelongings = 0
14442444
3
ΣFy
(4.9b)
Therefore, the magnitude of the normal force is FN = W + Wbelongings .The work done by the
normal force is
W = ( FN cos θ ) s = (W + Wbelongings ) ( cos θ ) s
(6.1)
= ( 685N + 915N )( cos0° ) (15.2m ) = 2 4 3 0 0 J
b. During the downward trip, you are still in equilibrium since the elevator is moving with a constant
velocity. The magnitude of the normal force is now FN = W. The work done by the normal force
is
W = ( FN cos θ ) s = ( W cos θ ) s = ( 685N )( cos180° )(15.2m ) = −1 0 4 0 0 J
(6.1)
______________________________________________________________________________
81. CONCEPT QUESTIONS
a. Since the asteroid is slowing down, it is decelerating. The force must point in a direction that is
opposite to that of the displacement.
b. The direction of the force is opposite to the displacement of the asteroid, so the angle between
them is 180°. Therefore, the force does negative work.
c. The asteroid is slowing down, so its speed is decreasing. Thus, its kinetic energy is changing.
d. The work done by the force is related to the change in the asteroid’s kinetic energy by the
work-energy theorem: W = 12 mv 2f − 12 mv02 , Equation 6.3.
SOLUTION
a. The work-energy theorem states that the work done by the force acting on the asteroid is equal
to its final kinetic energy minus its initial kinetic energy:
Chapter 6 Problems
W = 12 mvf2 −
=
1
2
1
2
249
mv02
(6.3)
( 4.5 × 10 kg ) ( 5500m/s)
4
2
−
1
2
( 4.5 × 10 kg ) ( 7100m/s)
4
2
= −4.5 × 1011 J
b. The work done by the force acting on the asteroid is given by Equation 6.1 as
W = ( F cos θ ) s . Since the force acts to slow down the asteroid, the force and the displacement
vectors point in opposite directions, so the angle between them is θ = 180°. The magnitude of the
force is
F=
W
−4.5 × 1011 J
=
= 2.5 × 105 N
6
s cosθ
1.8 × 10 m cos180°
(
)
(6.1)
______________________________________________________________________________
82. CONCEPT QUESTIONS
a. Since there is no friction, there are only two forces acting on each box, the gravitational force
(its weight) and the normal force. The normal force, being perpendicular to the displacement of a
box, does no work. The gravitational force does work. However, it is a conservative force, so the
conservation of mechanical energy applies to the motion of each box. This principle states that the
final speed depends only on the initial speed, and the initial and final heights. The final speed does
not depend on the mass of the box or on the steepness of the slope. Thus, both boxes reach B
with the same speed.
b. The kinetic energy is given by the expression K E = 12 mv2 . Both boxes have the same speed v
when they reach B. However, the heavier box has the greater mass. Therefore, the heavier box has
the greater kinetic energy.
SOLUTION
a. The conservation of mechanical energy states that
1 mv 2
B
2
+ mghB =
1 mv 2
A
2
+ mghA
(6.9b)
Solving for the speed v B at B gives
vB =
vA2 + 2 g ( hA − hB ) =
( 0 m /s )
2
(
)
+ 2 9.80m/s 2 ( 4.5m − 1.5m ) =
b. The speed of the heavier box is the same as the lighter box: 7.7m/s .
c. The ratio of the kinetic energies is
7.7m/s
250 WORK AND ENERGY
( KE B ) heavierbox
( KE B ) lighterbox
(
=
(
)
mv )
1
2
mvB2
1
2
2
B
heavier box
=
mheavier box
mlighterbox
=
44kg
=
11kg
4.0
lighterbox
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83. CONCEPT QUESTIONS
a. The kinetic frictional force does negative work, because the force is directed opposite to the
displacement of the student.
b. Since the kinetic frictional force does negative work, the final total mechanical energy must be
less than the initial total mechanical energy. Therefore, the total mechanical energy decreases.
c. The work done by the kinetic frictional force is related to the change in the student’s total
mechanical energy by the work-energy theorem, Wnc = E f − E 0 , Equation 6.8.
SOLUTION
The work-energy theorem states that
(
)
Wnc = 12 mv f2 + mghf − 12 mv02 + mgh0
144244
3
144
42444
3
Ef
E0
Solving for the final speed gives
vf =
=
(6.8)
2Wnc
+ v02 + 2 g ( h0 − hf )
m
(
2 −6.50 × 103 J
) + (0 m / s )
2
(
)
+ 2 9.80m/s 2 (11.8m ) = 8 . 6 m / s
83.0kg
______________________________________________________________________________
84. CONCEPT QUESTIONS
a. The two types of energy that are changing are the kinetic energy and the gravitational potential
energy. The kinetic energy is increasing, because the speed of the helicopter is increasing. The
gravitational potential energy is increasing, because the height of the helicopter is increasing.
b. The work done by the lifting force is related to the kinetic and potential energies by the workenergy theorem, Equation 6.6.
c. The average power is defined as the work divided by the time, Equation 6.10a, so the time must
be known in addition to the work.
SOLUTION
Chapter 6 Problems
251
The average power generated by the lifting force is equal to the work done by the force divided by
the elapsed time (Equation 6.10a). The work can be related to the change in the helicopter’s total
mechanical energy, which can be determined. The average power is
P=
Wnc
t
(6.10a)
where Wnc is the work done by the nonconservative lifting force. This work is related to the helicopter’s
kinetic and potential energies by Equation 6.6, Wnc =
(
1
2
)
mvf2 − 12 mv02 + ( mghf − mgh0 ) , so the
average power is
W
P = nc =
t
(
1 mv 2
f
2
− 12 mv02
) + (mgh
f
− mgh0 )
t
(
=
1m
2
(v
2
f
− v02
) + mg ( h
f
− h0 )
t
)
2
− ( 0 m /s )  + ( 810kg ) 9.80m/s 2 [8.2m − 0 m ]
P=
= 2.4 × 104 W
3.5s
______________________________________________________________________________
1
2
( 810kg ) ( 7.0m/s )
2
85. CONCEPT QUESTIONS
a. The kinetic energy is changing because the skier’s speed in changing.
b. The work is positive. Because the speed of the skier is increasing, the net external force must be
in the same direction as the displacement of the skier.
c. The work done by the net external force is equal to the change in the kinetic energy of the skier.
SOLUTION The work done by the net external force acting on the skier can be found from the
work-energy theorem in the form of Equation 6.3, because the final and initial kinetic energies of
the skier can be determined:
W = 12 mvf2 −
=
1
2
1
2
mv02
( 70.3kg ) (11.3m/s ) 2 − 12 ( 70.3kg )( 6.10m/s )2
= 3.2 × 10 3 J
______________________________________________________________________________