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

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°

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

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

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

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.

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

gv

S2

32.17 10

0.52

Problem 15.51 In Problem 15.50, what spring con-

stant is necessary if the coefﬁcient of kinetic friction

between the package and the ﬂoor is µk=0.3 and the

package contacts the bumper moving at 10 ft/s?

202

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 coefﬁcient 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

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

f−0

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

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 coefﬁcient

of kinetic friction between the 4-kg mass and the hori-

zontal surface is µk=0.4.

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 coefﬁcient

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

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:

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

2−1

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

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

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

Problem 15.67 A spring-powered mortar is used to

launch 10-lb packages of ﬁreworks 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

1−r2).

θ

r

Solution: The work done on the object is

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=(6−1)2+(2−1)2+(1−0)2

212

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)2−33 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)2−33 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:

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

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 ﬁeld is F=qv×B, where qand vare the

charge and velocity of the particle and Bis the magnetic

ﬁeld 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.

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 ﬂoor 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

(a) (b)

Solution:

216

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

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(m1−m2)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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