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Problem 15.1 In Active Example 15.1, what is the
velocity of the container when it has reached the position
s=2m?
s
A
Problem 15.2 The mass of the Sikorsky UH-60A heli-
copter is 9300 kg. It takes off vertically with its rotor
exerting a constant upward thrust of 112 kN. Use the
principle of work and energy to determine how far it
has risen when its velocity is 6 m/s.
Strategy: Be sure to draw the free-body diagram of
the helicopter.
Problem 15.3 The 20-lb box is at rest on the horizon-
tal surface when the constant force F=5 lb is applied.
The coefficient of kinetic friction between the box and
the surface is µk=0.2. Determine how fast the box is
moving when it has moved 2 ft from its initial position
(a) by applying Newton’s second law; (b) by applying
the principle of work and energy.
F
Solution:
Problem 15.4 At the instant shown, the 30-lb box is
moving up the smooth inclined surface at 2 ft/s. The con-
stant force F=15 lb. How fast will the box be moving
when it has moved 1 ft up the surface from its present
position?
F
20⬚
Solution:
180
Problem 15.5 The 0.45-kg soccer ball is 1 m above
the ground when it is kicked straight upward at 10 m/s.
By using the principle of work and energy, determine:
(a) how high above the ground the ball goes, (b) the
magnitude of the ball’s velocity when it falls back to a
height of 1 m above the ground, (c) the magnitude of
the ball’s velocity immediately before it hits the ground.
12 m/s
1 m
Solution:
Problem 15.6 Assume that the soccer ball in Prob-
Solution:
Problem 15.7 The 2000-lb drag racer starts from rest
and travels a quarter-mile course. It completes the course
in 4.524 seconds and crosses the finish line traveling at
325.77 mi/h. (a) How much work is done on the car as it
travels the course? (b) Assume that the horizontal force
exerted on the car is constant and use the principle of
work and energy to determine it.
Solution:
Problem 15.8 The 2000-lb drag racer starts from rest
and travels a quarter-mile course. It completes the course
in 4.524 seconds and crosses the finish line traveling at
325.77 mi/h. Assume that the horizontal force exerted on
the car is constant. Determine (a) the maximum power
and (b) the average power transferred to the car as it
travels the quarter-mile course.
182
Problem 15.9 As the 32,000-lb airplane takes off, the
tangential component of force exerted on it by its engines
is Ft=45,000 lb. Neglecting other forces on the air-
plane, use the principle of work and energy to determine
how much runway is required for its velocity to reach
200 mi/h.
Solution:
Problem 15.10 As the 32,000-lb airplane takes off, the
tangential component of force exerted on it by its engines
is Ft=45,000 lb. Neglecting other forces on the air-
plane, determine (a) the maximum power and (b) the
average power transferred to the airplane as its velocity
increases from zero to 200 mi/h.
Solution:
Problem 15.11 The 32,000-lb airplane takes off from
rest in the position s=0. The total tangential force
exerted on it by its engines and aerodynamic drag (in
pounds) is given as a function of its position sby Ft=
45,000 −5.2s. Use the principle of work and energy to
determine how fast the airplane is traveling when its
position is s=950 ft.
Solution:
Problem 15.12 The spring (k=20 N/m) is un-
stretched when s=0. The 5-kg cart is moved to the
position s=−1 m and released from rest. What is the
magnitude of its velocity when it is in the position s=0?
s
k
20⬚
Solution: First we calculate the work done by the spring and by
184
Problem 15.13 The spring (k=20 N/m) is un-
stretched when s=0. The 5-kg cart is moved to the
position s=−1 m and released from rest. What max-
imum distance down the sloped surface does the cart
move relative to its initial position?
s
k
20⬚
Solution: The cart starts from a position of rest, and when it
Problem 15.14 The force exerted on a car by a proto-
type crash barrier as the barrier crushes is F=−(120s +
40s3)lb, where sis the distance in feet from the initial
contact. The effective length of the barrier is 18 ft. How
fast can a 5000-lb car be moving and be brought to rest
within the effective length of the barrier?
s
Solution: The barrier can provide a maximum amount of work
given by
Problem 15.15 A 5000-lb car hits the crash barrier at
80 mi/h and is brought to rest in 0.11 seconds. What
average power is transferred from the car during the
impact?
s
Problem 15.16 A group of engineering students con-
structs a sun-powered car and tests it on a circular track
with a 1000-ft radius. The car, with a weight of 460 lb
including its occupant, starts from rest. The total tangen-
tial component of force on the car is
Ft=30 −0.2slb,
where sis the distance (in ft) the car travels along the
track from the position where it starts.
(a) Determine the work done on the car when it has
gone a distance s=120 ft.
(b) Determine the magnitude of the total horizontal
force exerted on the car’s tires by the road when it
is at the position s=120 ft.
Solution:
186
Problem 15.17 At the instant shown, the 160-lb
vaulter’s center of mass is 8.5 ft above the ground, and
the vertical component of his velocity is 4 ft/s. As his
pole straightens, it exerts a vertical force on the vaulter
of magnitude 180 +2.8y2lb, where yis the vertical
position of his center of mass relative to its position at
the instant shown. This force is exerted on him from
y=0toy=4 ft, when he releases the pole. What is
the maximum height above the ground reached by the
vaulter’s center of mass?
Problem 15.18 The springs (k=25 lb/ft) are un-
stretched when s=0. The 50-lb weight is released from
rest in the position s=0.
(a) When the weight has fallen 1 ft, how much work
has been done on it by each spring?
(b) What is the magnitude of the velocity of the weight
when it has fallen 1 ft? s
k
k
Problem 15.19 The coefficients of friction between
the 160-kg crate and the ramp are µs=0.3 and
µk=0.28.
(a) What tension T0must the winch exert to start the
crate moving up the ramp?
(b) If the tension remains at the value T0after the crate
starts sliding, what total work is done on the crate
as it slides a distance s=3 m up the ramp, and
what is the resulting velocity of the crate?
18°s
(b) The work done on the crate by (non-friction) external forces is
Uweight =3
0
T0ds −3
0
(mg sin θ)ds =932.9(3)−1455.1
=1343.5 N-m.
The work done on the crate by friction is
Uf=3
0
(−µkN)ds =−3µkmg cos θ=−1253.9 N-m.
From the principle of work and energy is
Uweight +Uf=1
2mv2,
from which
v=6(T0−mg(sin θ+µkcos θ))
m
v=1.06 m/s
Problem 15.20 In Problem 15.19, if the winch exerts
a tension T=T0(1+0.1s) after the crate starts sliding,
what total work is done on the crate as it slides a distance
s=3 m up the ramp, and what is the resulting velocity
of the crate?
0
0
0
from which
U=T0s+0.05s23
0−(mg sin θ)(3)−µk(mg cos θ)(3).
From the solution to Problem 15.19, T0=932.9 N-m, from which the
total work done is
U=3218.4−1455.1−1253.9=509.36 N-m.
v=2U
m=2.52 m/s
188
c
2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 15.21 The 200-mm-diameter gas gun is
evacuated on the right of the 8-kg projectile. On
the left of the projectile, the tube contains gas
with pressure p0=1×105Pa (N/m2). The force F
is slowly increased, moving the projectile 0.5 m to
the left from the position shown. The force is
then removed and the projectile accelerates to the
right. If you neglect friction and assume that the
pressure of the gas is related to its volume by
pV =constant, what is the velocity of the projectile
when it has returned to its original position?
Projectile
F
Gas
1 m
Problem 15.22 In Problem 15.21, if you assume that
the pressure of the gas is related to its volume by
pV =constant while it is compressed (an isothermal
process) and by pV 1.4=constant while it is expand-
ing (an isentropic process), what is the velocity of the
projectile when it has returned to its original position?
Solution: The isothermal constant is K=3141.6 N-m from the
solution to Problem 15.21. The pressure at the leftmost position is
The pressure during expansion is
From the principle of work and energy, the work done is equal to the
gain in kinetic energy,
m=21.8 m/s .
Problem 15.23 In Example 15.2, suppose that the angle
between the inclined surface and the horizontal is in-
creased from 20◦to 30◦. What is the magnitude of the
velocity of the crates when they have moved 400 mm?
v
A
20⬚
2(70)v2
Solving for the velocity we find
Problem 15.24 The system is released from rest. The
4-kg mass slides on the smooth horizontal surface. By
using the principle of work and energy, determine the
magnitude of the velocity of the masses when the 20-kg
mass has fallen 1 m.
20 kg
4 kg
Problem 15.25 Solve Problem 15.24 if the coefficient
of kinetic friction between the 4-kg mass and the hori-
zontal surface is µk=0.4.
Solution:
U=[(20 kg)(9.81 m/s2)−0.4(39.24 N)](1m)=180.5 N-m
180.5 N-m =1
2(24 kg)v2⇒v=3.88 m/s
N
(4 kg)(9.81 m/s2)
190
c
2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 15.26 Each box weighs 50 lb and the
inclined surfaces are smooth. The system is released
from rest. Determine the magnitude of the velocities of
the boxes when they have moved 1 ft.
45°
30°
Problem 15.27 Solve Problem 15.26 if the coefficient
of kinetic friction between the boxes and the inclined
surfaces is µk=0.05.
Solution:
50 lb 50 lb
Problem 15.28 The masses of the three blocks are
mA=40 kg, mB=16 kg, and mC=12 kg. Neglect the
mass of the bar holding Cin place. Friction is negligible.
By applying the principle of work and energy to Aand B
individually, determine the magnitude of their velocity
when they have moved 500 mm.
A
B
C
45⬚
0
2mBv2.
Add the two equations:
(mA−mB)gb sin θ=1
2(mA+mB)v2.
Solve: |vA|=|vB|=2(mA−mB)gb sin θ
(mA+mB)=1.72 m/s
NA
Problem 15.29 Solve Problem 15.28 by applying the
principle of work and energy to the system consisting of
A,B, the cable connecting them, and the pulley.
Solution: Choose a coordinate system with the origin at the pulley
192
Problem 15.30 The masses of the three blocks are
mA=40 kg, mB=16 kg, and mC=12 kg. The coeffi-
cient of kinetic friction between all surfaces is µk=0.1.
Determine the magnitude of the velocity of blocks Aand
Bwhen they have moved 500 mm. (See Example 15.3.)
A
B
C
Solution: We will apply the principles of work — energy to blocks
Aand Bindividually in order to properly account for the work done
by internal friction forces.
b
(mAgsin θ−T−µkNA−µkNAB )ds=1
Problem 15.31 In Example 15.5, suppose that the skier
is moving at 20 m/s when he is in position 1. Deter-
mine the horizontal component of his velocity when he
reaches position 2, 20 m below position 1.
1
2
3
Problem 15.32 Suppose that you stand at the edge of
a 200-ft cliff and throw rocks at 30 ft/s in the three
directions shown. Neglecting aerodynamic drag, use the
principle of work and energy to determine the magnitude
of the velocity of the rock just before it hits the ground
in each case.
30⬚
30⬚
200 ft
(a)
(b)
(c)
Problem 15.33 The 30-kg box is sliding down the
smooth surface at 1 m/s when it is in position 1.
Determine the magnitude of the box’s velocity at
position 2 in each case.
1
1
2 m
Problem 15.34 Solve Problem 15.33 if the coefficient
of kinetic friction between the box and the inclined sur-
Solution: The work done by the weight is the same, however, the
work done by friction is different.
Problem 15.35 In case (a), a 5-lb ball is released from
rest at position 1 and falls to position 2. In case (b), the
ball is released from rest at position 1 and swings to
position 2. For each case, use the principle of work and
energy to determine the magnitude of the ball’s velocity
at position 2. (In case (b), notice that the force exerted on
the ball by the string is perpendicular to the ball’s path.)
2 ft
11
194
Problem 15.36 The 2-kg ball is released from rest in
position 1 with the string horizontal. The length of the
string is L=1 m. What is the magnitude of the ball’s
velocity when it is in position 2? L
2
1
40⬚
Solution:
2 kg
1
L = 1 m
Problem 15.37 The 2-kg ball is released from rest in
position 1 with the string horizontal. The length of the
string is L=1 m. What is the tension in the string when
the ball is in position 2?
Strategy: Draw the free-body diagram of the ball
when it is in position 2 and write Newton’s second law
in terms of normal and tangential components.
Problem 15.38 The 400-lb wrecker’s ball swings at
the end of a 25-ft cable. If the magnitude of the ball’s
velocity at position 1 is 4 ft/s, what is the magnitude of
its velocity just before it hits the wall at position 2?
12
95°
65°
Solution:
Problem 15.39 The 400-lb wrecker’s ball swings at
the end of a 25-ft cable. If the magnitude of the ball’s
velocity at position 1 is 4 ft/s, what is the maximum
tension in the cable as the ball swings from position 1
to position 2?
196
Problem 15.40 A stunt driver wants to drive a car
through the circular loop of radius R=5 m. Determine
the minimum velocity v0at which the car can enter the
loop and coast through without losing contact with the
track. What is the car’s velocity at the top of the loop?
R
0
Solution: First, let us find VT
VTOP = VT
Problem 15.41 The 2-kg collar starts from rest at posi-
tion 1 and slides down the smooth rigid wire. The y-axis
points upward. What is the magnitude of the velocity of
the collar when it reaches position 2?
(3, –1, 3) m
x
y
z
(5, 5, 2) m1
2
2 kg
Problem 15.42 The 4-lb collar slides down the smooth
rigid wire from position 1 to position 2. When it reaches
position 2, the magnitude of its velocity is 24 ft/s. What
was the magnitude of its velocity at position 1?
(4, ⫺1, 4) ft
x
y
z
(⫺2, 6, 4) ft
1
2
4 lb
Problem 15.43 The forces acting on the 28,000-lb air-
plane are the thrust T and drag D, which are parallel to
the airplane’s path, the lift L, which is perpendicular to
the path, and the weight W. The airplane climbs from
an altitude of 3000 ft to an altitude of 10,000 ft. During
the climb, the magnitude of its velocity decreases from
800 ft/s to 600 ft/s.
(a) What work is done on the airplane by its lift during
the climb?
(b) What work is done by the thrust and drag com-
bined?
T
W
D
L
Solution:
198
