Mechanical Engineering Chapter 15 Problem The Car Traveling Mih Position The Combined Effect The Aerodynamicdrag The

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subject Authors Anthony M. Bedford, Wallace Fowler

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Problem 15.44 The 2400-lb car is traveling 40 mi/h
at position 1. If the combined effect of the aerodynamic
drag on the car and the tangential force exerted on its
wheels by the road is that they exert no net tangential
force on the car, what is the magnitude of the car’s
velocity at position 2?
30°
30°
120 ft
100 ft
1
2
Solution: The initial velocity is
0
From the principle of work and energy the work done is equal to the
Problem 15.45 The 2400-lb car is traveling 40 mi/h at
position 1. If the combined effect of aerodynamic drag
on the car and the tangential force exerted on its wheels
by the road is that they exert a constant 400-lb tangential
force on the car in the direction of its motion, what is
the magnitude of the car’s velocity at position 2?
Utgt =s
0
400 ds =400(115.2)=46076.7 ft-lb.
=52.73 ft/s =36 mph
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Problem 15.46 The mass of the rocket is 250 kg. Its
engine has a constant thrust of 45 kN. The length of the
launching ramp is 10 m. If the magnitude of the rocket’s
velocity when it reaches the end of the ramp is 52 m/s,
how much work is done on the rocket by friction and
aerodynamic drag?
2 m
Problem 15.47 A bioengineer interested in energy
requirements of sports determines from videotape that
when the athlete begins his motion to throw the 7.25-kg
shot (Fig. a), the shot is stationary and 1.50 m above
the ground. At the instant the athlete releases it (Fig. b),
the shot is 2.10 m above the ground. The shot reaches a
maximum height of 4.60 m above the ground and travels
a horizontal distance of 18.66 m from the point where
it was released. How much work does the athlete do on
the shot from the beginning of his motion to the instant
he releases it?
200
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Problem 15.48 A small pellet of mass m=0.2kg
starts from rest at position 1 and slides down the smooth
surface of the cylinder to position 2, where θ=30.
(a) What work is done on the pellet as it slides from
position 1 to position 2?
(b) What is the magnitude of the pellet’s velocity at
position 2?
1
2
0.8 m
m
u
=0.210 N-m.
(a) U12 =0.210 (N-m)
(b) 0.1v2
2=0.210 (N-m)
v2=1.45 m/s
Problem 15.49 In Active Example 15.4, suppose that
you want to increase the value of the spring constant k
so that the velocity of the hammer just before it strikes
the workpiece is 4 m/s. what is the required value of k?
400
m
m
2
1
kk
Hammer
Solution: The 40-kg hammer is released from rest in position 1.
The springs are unstretched when in position 2. Neglect friction.
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Problem 15.50 Suppose that you want to design a
bumper that will bring a 50-lb. package moving at 10 ft/s
to rest 6 in from the point of contact with bumper.
If friction is negligible, what is the necessary spring
constant k?k
10 ft/s
gv
S2
32.17 10
0.52
Problem 15.51 In Problem 15.50, what spring con-
stant is necessary if the coefcient of kinetic friction
between the package and the oor is µk=0.3 and the
package contacts the bumper moving at 10 ft/s?
202
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Problem 15.52 The 50-lb package starts from rest,
slides down the smooth ramp, and is stopped by the
spring.
(a) If you want the package to be brought to rest 6
in from the point of contact, what is the necessary
spring constant k?
(b) What maximum deceleration is the package sub-
jected to?
k
4 ft
32.2 ft/s2(0.5ft)(32.2 ft/s2)sin 30
Problem 15.53 The 50-lb package starts from rest,
slides down the smooth ramp, and is stopped by the
spring. The coefcient of static friction between the pack-
age and the ramp is µk=0.12. If you want the package
to be brought to rest 6 in from the point of contact, what
is the necessary spring constant k?
k
4 ft
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Problem 15.54 The system is released from rest
with the spring unstretched. The spring constant
k=200 N/m. Determine the magnitude of the velocity
of the masses when the right mass has fallen 1 m.
k
20 kg
4 kg
Solution: When the larger mass falls 1 m, the smaller mass rises
U12 =56.96 N-m
Also U12 =1
2(m1+m2)V 2
f0
Solving Vf=2.18 m/s
k
Problem 15.55 The system is released from rest
with the spring unstretched. The spring constant
k=200 N/m. What maximum downward velocity does
the right mass attain as it falls?
For all s. Setting these equal, we get
K=0.785 m
204
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Problem 15.56 The system is released from rest. The
4-kg mass slides on the smooth horizontal surface. The
spring constant is k=100 N/m, and the tension in the
spring when the system is released is 50 N. 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
k
Solution:
Problem 15.57 Solve Problem 15.56 if the coefcient
of kinetic friction between the 4-kg mass and the hori-
zontal surface is µk=0.4.
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Problem 15.58 The 40-lb crate is released from rest on
the smooth inclined surface with the spring unstretched.
The spring constant is k=8 lb/ft.
(a) How far down the inclined surface does the crate
slide before it stops?
(b) What maximum velocity does the crate attain on
its way down?
k
30
Solution: At an arbitrary distance s down the slope we have:
Problem 15.59 Solve Problem 15.58 if the coefcient
of kinetic friction between the 4-kg mass and the hori-
zontal surface is µk=0.2.
Solution: At an arbitrary distance s down the slope we have:
206
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Problem 15.60 The 4-kg collar starts from rest in posi-
tion 1 with the spring unstretched. The spring constant
is k=100 N/m. How far does the collar fall relative to
position 1?
k
1
Solution:
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Problem 15.61 In position 1 on the smooth bar, the
4-kg collar has a downward velocity of 1 m/s and
the spring is unstretched. The spring constant is k=
100 N/m. What maximum downward velocity does the
collar attain as it falls?
Thus,
1
2mV 2
21
2mV 2
1=−Ks2
2+mgs (1)
Finding dV2
ds , and setting it to zero,
mV2
dV2
ds =−Ks +mg =0
s=mg/ k =0.392 m
Solving (1) for V2we get V2=2.20 m/s
Problem 15.62 The 4-kg collar starts from rest in posi-
tion 1 on the smooth bar. The tension in the spring in
position 1 is 20 N. The spring constant is k=100 N/m.
How far does the collar fall relative to position 1?
208
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Problem 15.63 The 4-kg collar is released from rest
at position 1 on the smooth bar. If the spring constant is
k=6 kN/m and the spring is unstretched in position 2,
what is the velocity of the collar when it has fallen to
position 2?
k
1
250 mm
Solution: Denote d=200 mm, h=250 mm. The stretch of the
Problem 15.64 The 4-kg collar is released from rest
in position 1 on the smooth bar. The spring constant is
k=4 kN/m. The tension in the spring in position 2 is
500 N. What is the velocity of the collar when it has
fallen to position 2?
Solution: Denote d=200 mm, h=250 mm. The stretch of the
page-pfc
Problem 15.65 The 4-kg collar starts from rest in
position 1 on the smooth bar. Its velocity when it has
fallen to position 2 is 4 m/s. The spring is unstretched
when the collar is in position 2. What is the spring
constant k?
Problem 15.66 The 10-kg collar starts from rest at
position 1 and slides along the smooth bar. The y-axis
points upward. The spring constant is k=100 N/m and
the unstretched length of the spring is 2 m. What is the
velocity of the collar when it reaches position 2?
y(4, 4, 2) m
(6, 2, 1) m
(1, 1, 0) m
1
2
Solution: The stretch of the spring at position 1 is
210
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Problem 15.67 A spring-powered mortar is used to
launch 10-lb packages of reworks into the air. The
package starts from rest with the spring compressed to
a length of 6 in. The unstretched length of the spring is
30 in. If the spring constant is k=1300 lb/ft, what is
the magnitude of the velocity of the package as it leaves
the mortar? 30 in
Problem 15.68 Suppose that you want to design the
mortar in Problem 15.67 to throw the package to a height
Solution: See the solution of Problem 15.67. Let v2be the
velocity as the package leaves the barrel. To reach 150 ft, mg(150
Problem 15.69 Suppose an object has a string or cable
with constant tension Tattached as shown. The force
exerted on the object can be expressed in terms of polar
coordinates as F=−Ter. Show that the work done on
the object as it moves along an arbitrary plane path
from a radial position r1to a radial position r2is U12 =
T(r
1r2).
θ
r
Solution: The work done on the object is
page-pfe
Problem 15.70 The 2-kg collar is initially at rest at
position 1. A constant 100-N force is applied to the
string, causing the collar to slide up the smooth vertical
bar. What is the velocity of the collar when it reaches
position 2? (See Problem 15.69.)
2
200 mm
500 mm
Solution: The constant force on the end of the string acts through
Problem 15.71 The 10-kg collar starts from rest at
position 1. The tension in the string is 200 N, and the
yaxis points upward. If friction is negligible, what is the
magnitude of the velocity of the collar when it reaches
position 2? (See Problem 15.69.)
y
(4, 4, 2) m
2
s=(61)2+(21)2+(10)2
212
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Problem 15.72 As the F/A-18 lands at 210 ft/s, the
cable from Ato Bengages the airplane’s arresting hook
at C. The arresting mechanism maintains the tension
in the cable at a constant value, bringing the 26,000-lb
airplane to rest at a distance of 72 ft. What is the tension
in the cable? (See Problem 15.69.)
72 ft
Solution:
(72 ft)2+(33 ft)233 ft)
Problem 15.73 If the airplane in Problem 15.72 lands
at 240 ft/s, what distance does it roll before the arresting
Solution:
U=−2(193,000 lb)(s2+(33 ft)233 ft)
Problem 15.74 A spacecraft 320 km above the sur-
face of the earth is moving at escape velocity vesc =
10,900 m/s. What is its distance from the center of the
earth when its velocity is 50 percent of its initial value?
The radius of the earth is 6370 km. (See Example 15.6.)
320 km
vesc
Solution:
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Problem 15.75 A piece of ejecta is thrown up by the
impact of a meteor on the moon. When it is 1000 km
above the moon’s surface, the magnitude of its velocity
(relative to a nonrotating reference frame with its ori-
gin at the center of the moon) is 200 m/s. What is the
magnitude of its velocity just before it strikes the moon’s
surface? The acceleration due to gravity at the surface of
the moon is 1.62 m/s2. The moon’s radius is 1738 km.
1000 km 200 m/s
Solution: The kinetic energy at h=1000 km is
214
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Problem 15.76 A satellite in a circular orbit of
radius raround the earth has velocity v=gR2
E/r,
where RE=6370 km is the radius of the earth. Suppose
you are designing a rocket to transfer a 900-kg
communication satellite from a circular parking orbit
with 6700-km radius to a circular geosynchronous orbit
with 42,222-km radius. How much work must the rocket
do on the satellite?
Problem 15.77 The force exerted on a charged particle
by a magnetic eld is F=qv×B, where qand vare the
charge and velocity of the particle and Bis the magnetic
eld vector. Suppose that other forces on the particle are
negligible. Use the principle of work and energy to show
that the magnitude of the particle’s velocity is constant.
page-pf12
Problem 15.78 The 10-lb box is released from rest at
position 1 and slides down the smooth inclined surface
to position 2.
(a) If the datum is placed at the level of the oor as
shown, what is the sum of the kinetic and potential
energies of the box when it is in position 1?
(b) What is the sum of the kinetic and potential ener-
gies of the box when it is in position 2?
(c) Use conservation of energy to determine the mag-
nitude of the box’s velocity when it is in position 2.
30
1
25 ft
2 ft
Datum
Solution:
Problem 15.79 The 0.45-kg soccer ball is 1 m above
the ground when it is kicked upward at 12 m/s. Use
conservation of energy to determine the magnitude of
the ball’s velocity when it is 4 m above the ground.
Obtain the answer by placing the datum (a) at the level
of the ball’s initial position and (b) at ground level.
12 m/s
1 m
12 m/s
1 m
Datum
Datum
(a) (b)
Solution:
216
page-pf13
Problem 15.80 The Lunar Module (LM) used in the
Apollo moon landings could make a safe landing if
the magnitude of its vertical velocity at impact was no
greater than 5 m/s. Use conservation of energy to deter-
mine the maximum height hat which the pilot could
shut off the engine if the vertical velocity of the lander
is (a) 2 m/s downward and (b) 2 m/s upward. The accel-
eration due to gravity at the moon’s surface is 1.62 m/s2.
h
Solution: Use conservation of energy Let state 1 be at the max
Problem 15.81 The 0.4-kg collar starts from rest at
position 1 and slides down the smooth rigid wire. The
yaxis points upward. Use conservation of energy to
determine the magnitude of the velocity of the collar
when it reaches point 2.
(3, 0,2) m
x
y
z
(5, 5, 2) m1
2
0.4 kg
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Problem 15.82 At the instant shown, the 20-kg mass
is moving downward at 1.6 m/s. Let dbe the downward
displacement of the mass relative to its present position.
Use conservation of energy to determine the magnitude
of the velocity of the 20-kg mass when d=1m.
Solution:
(m1+m2)v2
1=(m1+m2)v2
2+2(m1m2)gd
Substituting known values and solving v2=3.95 m/s
Problem 15.83 The mass of the ball is m=2 kg and
the string’s length is L=1 m. The ball is released from
rest in position 1 and swings to position 2, where θ=
40.
(a) Use conservation of energy to determine the mag-
nitude of the ball’s velocity at position 2.
(b) Draw graphs of the kinetic energy, the potential
energy, and the total energy for values of θfrom
zero to 180.
L
m2
1
u
Solution:
Total energy is always zero (datum value).
(a) Evaluating at θ=40,v2=3.55 m/s
Kinetic and Potential Energy vs
α
218
c
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