Dynamics 14th edition by r c hibbeler chapter 14

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14–1.
The 20-kg crate is subjected to a force having a constant
direction and a magnitude F = 100 N. When s = 15 m, the
crate is moving to the right with a speed of 8 m/s. Determine
its speed when s = 25 m. The coefficient of kinetic friction
between the crate and the ground is mk = 0.25.

F
30°

SOLUTION
Equation of Motion: Since the crate slides, the friction force developed between the
crate and its contact surface is Ff = mkN = 0.25N. Applying Eq. 13–7, we have
+ c a Fy = may ;

N + 100 sin 30° - 20(9.81) = 20(0)
N = 146.2 N

Principle of Work and Energy: The horizontal component of force F which acts
in the direction of displacement does positive work, whereas the friction force
Ff = 0.25(146.2) = 36.55 N does negative work since it acts in the opposite direction
to that of displacement. The normal reaction N, the vertical component of force F
and the weight of the crate do not displace hence do no work. Applying Eq.14–7,
we have
T1 + a U1 - 2 = T2
25 m

1

(20)(8 2) +
2
L15 m

100 cos 30° ds

25 m

-

L15 m

36.55 ds =

1
(20)v2
2
Ans.

v = 10.7 m s

Ans:
v = 10.7 m>s
377


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14–2.

F (lb)

For protection, the barrel barrier is placed in front of the
bridge pier. If the relation between the force and deflection
of the barrier is F = (90(103)x1>2) lb, where x is in ft,
determine the car’s maximum penetration in the barrier.
The car has a weight of 4000 lb and it is traveling with a
speed of 75 ft>s just before it hits the barrier.

F ϭ 90(10)3 x1/2

x (ft)

SOLUTION
Principle of Work and Energy: The speed of the car just before it crashes into the
barrier is v1 = 75 ft>s. The maximum penetration occurs when the car is brought to a
stop, i.e., v2 = 0. Referring to the free-body diagram of the car, Fig. a, W and N do no
work; however, Fb does negative work.
T1 + ©U1 - 2 = T2
1 4000
a
b (752) + c 2 32.2
L0

xmax

90(103)x1>2dx d = 0
Ans.

xmax = 3.24 ft

Ans:
x max = 3.24 ft
378


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14–3.
The crate, which has a mass of 100 kg, is subjected to the
action of the two forces. If it is originally at rest, determine
the distance it slides in order to attain a speed of 6 m>s. The
coefficient of kinetic friction between the crate and the
surface is mk = 0.2.

1000 N

800 N
5

30

3
4

SOLUTION
Equations of Motion: Since the crate slides, the friction force developed between
the crate and its contact surface is Ff = mk N = 0.2N. Applying Eq. 13–7, we have
+ c ©Fy = may;

3
N + 1000 a b - 800 sin 30° - 100(9.81) = 100(0)
5
N = 781 N

Principle of Work and Energy: The horizontal components of force 800 N and
1000 N which act in the direction of displacement do positive work, whereas the
friction force Ff = 0.2(781) = 156.2 N does negative work since it acts in the
opposite direction to that of displacement. The normal reaction N, the vertical
component of 800 N and 1000 N force and the weight of the crate do not displace,
hence they do no work. Since the crate is originally at rest, T1 = 0. Applying
Eq. 14–7, we have
T1 + a U1-2 = T2
4
1
0 + 800 cos 30°(s) + 1000 a bs - 156.2s = (100) A 62 B
5
2
Ans.

s = 1.35m

Ans:
s = 1.35 m
379


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*14–4.
The 100-kg crate is subjected to the forces shown. If it is
originally at rest, determine the distance it slides in order to
attain a speed of v = 8 m>s. The coefficient of kinetic
friction between the crate and the surface is mk = 0.2.

500 N

400 N
30Њ

45Њ

Solution
Work. Consider the force equilibrium along the y axis by referring to the FBD of
the crate, Fig. a,
+ c ΣFy = 0;

N + 500 sin 45° - 100(9.81) - 400 sin 30° = 0
N = 827.45 N

Thus, the friction is Ff = mkN = 0.2(827.45) = 165.49 N. Here, F1 and F2 do positive
work whereas Ff   does negative work. W and N do no work
UF1 = 400 cos 30° s = 346.41 s
UF2 = 500 cos 45° s = 353.55 s
UFf = - 165.49 s
Principle of Work And Energy. Applying Eq. 14–7,
T1 + ΣU1 - 2 = T2
0 + 346.41 s + 353.55 s + ( -165.49 s) =

1
(100) ( 82 )
2

s = 5.987 m = 5.99 m

Ans.

Ans:
s = 5.99 m
380


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14–5.
Determine the required height h of the roller coaster so that
when it is essentially at rest at the crest of the hill A it will
reach a speed of 100 km>h when it comes to the bottom B.
Also, what should be the minimum radius of curvature r for
the track at B so that the passengers do not experience a
normal force greater than 4mg = (39.24m) N? Neglect the
size of the car and passenger.

A

h

r
B

Solution
100 km>h =

100 ( 103 )
3600

= 27.778 m>s

T1 + ΣU1 - 2 = T2
0 + m(9.81)h =

1
m(27.778)2
2
Ans.

h = 39.3 m
+ c ΣFn = man;


(27.778)2
39.24 m - mg = ma
b
r

Ans.

r = 26.2 m

Ans:
h = 39.3 m
r = 26.2 m
381


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14–6.
When the driver applies the brakes of a light truck traveling
40 km>h, it skids 3 m before stopping. How far will the truck
skid if it is traveling 80 km>h when the brakes are applied?

Solution
40 km>h =

40 ( 103 )
3600

= 11.11 m>s

80 km>h = 22.22 m>s

T1 + ΣU1 - 2 = T2
1
m(11.11)2 - mk mg(3) = 0
2

mk g = 20.576
T1 + ΣU1 - 2 = T2
1
m(22.22)2 - (20.576)m(d) = 0
2
Ans.

d = 12 m

Ans:
d = 12 m
382


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14–7.
As indicated by the derivation, the principle of work and
energy is valid for observers in any inertial reference frame.
Show that this is so, by considering the 10-kg block which
rests on the smooth surface and is subjected to a horizontal
force of 6 N. If observer A is in a fixed frame x, determine the
final speed of the block if it has an initial speed of 5 m>s and
travels 10 m, both directed to the right and measured from
the fixed frame. Compare the result with that obtained by an
observer B, attached to the x¿ axis and moving at a constant
velocity of 2 m>s relative to A. Hint: The distance the block
travels will first have to be computed for observer B before
applying the principle of work and energy.

A
B

x
x¿

2 m/s
5 m/s
6N

10 m

SOLUTION
Observer A:
T1 + ©U1 - 2 = T2
1
1
(10)(5)2 + 6(10) = (10)v22
2
2
Ans.

v2 = 6.08 m>s
Observer B:
F = ma
6 = 10a
+ B
A:

a = 0.6 m>s2
s = s0 + v0t +

1 2
at
2 c

10 = 0 + 5t +

1
(0.6)t2
2

t2 + 16.67t - 33.33 = 0
t = 1.805 s
At v = 2 m>s, s¿ = 2(1.805) = 3.609 m
Block moves 10 - 3.609 = 6.391 m
Thus
T1 + ©U1 - 2 = T2
1
1
(10)(3)2 + 6(6.391) = (10)v22
2
2
Ans.

v2 = 4.08 m>s
Note that this result is 2 m>s less than that observed by A.

Ans:

Observer A: v2 = 6.08 m>s
Observer B: v2 = 4.08 m>s
383


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*14–8.
A force of F = 250 N is applied to the end at B. Determine
the speed of the 10-kg block when it has moved 1.5 m,
starting from rest.

Solution
Work. with reference to the datum set in Fig. a,

B

SW + 2sF = l
(1)

dSW + 2dsF = 0



A

F

Assuming that the block moves upward 1.5 m, then dSW = -1.5 m since it is directed

in the negative sense of SW. Substituted this value into Eq. (1),
- 1.5 + 2dsF = 0  dsF = 0.75 m
Thus,
UF = FdSF = 250(0.75) = 187.5 J
UW = - WdSW = - 10(9.81)(1.5) = -147.15 J
Principle of Work And Energy. Applying Eq. 14–7,
T1 + U1 - 2 = T2
0 + 187.5 + ( -147.15) =

1
(10)v2
2
Ans.

v = 2.841 m>s = 2.84 m>s

Ans:
v = 2.84 m>s
384


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14–9.
The “air spring” A is used to protect the support B and
prevent damage to the conveyor-belt tensioning weight C
in the event of a belt failure D. The force developed by
the air spring as a function of its deflection is shown by the
graph. If the block has a mass of 20 kg and is suspended

a height d = 0.4 m above the top of the spring, determine
the maximum deformation of the spring in the event the
conveyor belt fails. Neglect the mass of the pulley and belt.

F (N)

D

1500

C
d

0.2

s (m)

A
B

Solution
Work. Referring to the FBD of the tensioning weight, Fig. a, W does positive
work whereas force F does negative work. Here the weight displaces downward
SW = 0.4 + xmax where xmax is the maximum compression of the air spring. Thus
UW = 20(9.81) ( 0.4 + xmax ) = 196.2 ( 0.4 + xmax )
The work of F is equal to the area under the F-S graph shown shaded in Fig. b, Here
F
1500
   ; F = 7500xmax. Thus
=

xmax
0.2
1
UF = - (7500 xmax)(xmax) = - 3750x2max
2
Principle of Work And Energy. Since the block is at rest initially and is required
to stop momentarily when the spring is compressed to the maximum, T1 = T2 = 0.
Applying Eq. 14–7,
T1 + ΣU1 - 2 = T2
0 + 196.2(0.4 + xmax) +

( - 3750x2max ) = 0

3750x2max - 196.2xmax - 78.48 = 0


xmax = 0.1732 m = 0.173 m 6 0.2 m      (O.K!)

Ans.

Ans:
xmax = 0.173 m
385


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14–10.
v

The force F, acting in a constant direction on the 20-kg
block, has a magnitude which varies with the position s of
the block. Determine how far the block must slide before its
velocity becomes 15 m>s. When s = 0 the block is moving
to the right at v = 6 m>s. The coefficient of kinetic friction
between the block and surface is mk = 0.3.

F (N)

F

F ϭ 50s 1/2

Solution
Work. Consider the force equilibrium along y axis, by referring to the FBD of the
block, Fig. a,

s (m)

+ c ΣFy = 0 
;   N - 20(9.81) = 0   N = 196.2 N
Thus, the friction is Ff = mkN = 0.3(196.2) = 58.86 N. Here, force F does positive
work whereas friction Ff does negative work. The weight W and normal reaction N
do no work.
UF =

L

Fds =

L0

s

1

50s2 ds =

100 3
s2
3

UFf = - 58.86 s
Principle of Work And Energy. Applying Eq. 14–7,
T1 + ΣU1 - 2 = T2
1
100 3
1
(20)(62) +
s2 + ( - 58.86s) = (20)(152)
2
3
2
100 3
s2 - 58.86s - 1890 = 0
3
Solving numerically,
Ans.

s = 20.52 m = 20.5 m

Ans:
s = 20.5 m
386


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14–11.
The force of F = 50 N is applied to the cord when s = 2 m.
If the 6-kg collar is orginally at rest, determine its velocity at
s = 0. Neglect friction.

F

1.5 m
A

Solution

s

Work. Referring to the FBD of the collar, Fig. a, we notice that force F
does positive work but W and N do no work. Here, the displacement of F is
s = 222 + 1.52 - 1.5 = 1.00 m

UF = 50(1.00) = 50.0 J

Principle of Work And Energy. Applying Eq. 14–7,
T1 + ΣU1 - 2 = T2
0 + 50 =


1
(6)v2
2

v = 4.082 m>s = 4.08 m>s

Ans.

Ans:
v = 4.08 m>s
387


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*14–12.
Design considerations for the bumper B on the 5-Mg train
car require use of a nonlinear spring having the loaddeflection characteristics shown in the graph. Select the
proper value of k so that the maximum deflection of the
spring is limited to 0.2 m when the car, traveling at 4 m>s,
strikes the rigid stop. Neglect the mass of the car wheels.

F (N)

F

ks2

s (m)

SOLUTION
B

0.2

1
ks2 ds = 0
(5000)(4)2—
2
L0
40 000 - k

(0.2)3
= 0
3

k = 15.0 MN>m2

Ans.

Ans:
k = 15.0 MN>m2
388

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

14–13.
The 2-lb brick slides down a smooth roof, such that when it
is at A it has a velocity of 5 ft>s. Determine the speed of the
brick just before it leaves the surface at B, the distance d
from the wall to where it strikes the ground, and the speed
at which it hits the ground.

y
A

5 ft/s

15 ft

5

3
4

B

SOLUTION
TA + ©UA-B = TB
30 ft

2

2
1
1
a
b(5)2 + 2(15) = a
bv2B
2 32.2
2 32.2
Ans.

vB = 31.48 ft>s = 31.5 ft>s
+ b
a:

Ans.
d

s = s0 + v0t
4
d = 0 + 31.48 a b t
5

A+TB

s = s0 + v0t -

1
ac t2
2

1
3
30 = 0 + 31.48 a b t + (32.2)t2
5
2
16.1t2 + 18.888t - 30 = 0
Solving for the positive root,
t = 0.89916 s
4
d = 31.48a b (0.89916) = 22.6 ft
5

Ans.

TA + ©UA-C = TC
2
2
1
1
a
b(5)2 + 2(45) = a
bv2C
2 32.2
2 32.2
Ans.

vC = 54.1 ft>s

Ans:
vB = 31.5 ft>s

d = 22.6 ft
vC = 54.1 ft>s
389

x


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14–14.
Block A has a weight of 60 lb and block B has a weight of
10 lb. Determine the speed of block A after it moves 5 ft
down the plane, starting from rest. Neglect friction and the
mass of the cord and pulleys.

B
A
5

SOLUTION

3

4

2 sA + sB = l
2 ∆sA + ∆sB = 0
2vA + vB = 0
T1 + ΣU1 - 2 = T2

3
1 10
1 60
0 + 60 a b(5) - 10(10) = a
b vA2 + a
b(2vA)2
5
2 32.2
2 32.2

Ans.

vA = 7.18 ft>s

Ans:
vA = 7.18 ft>s
390


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

14–15.
The two blocks A and B have weights WA = 60 lb and
WB = 10 lb. If the kinetic coefficient of friction between the
incline and block A is mk = 0.2, determine the speed of A
after it moves 3 ft down the plane starting from rest. Neglect
the mass of the cord and pulleys.
A

SOLUTION

5

Kinematics: The speed of the block A and B can be related by using position
coordinate equation.
sA + (sA - sB) = l
2¢sA - ¢sB = 0

B

3
4

2sA - sB = l

¢sB = 2¢sA = 2(3) = 6 ft
(1)

2vA - vB = 0
Equation of Motion: Applying Eq. 13–7, we have
+ ©Fy¿ = may¿ ;

60
4
(0)
N - 60 a b =
5
32.2

N = 48.0 lb

Principle of Work and Energy: By considering the whole system, WA which acts in
the direction of the displacement does positive work. WB and the friction force
Ff = mkN = 0.2(48.0) = 9.60 lb does negative work since they act in the opposite
direction to that of displacement Here, WA is being displaced vertically (downward)
3
¢s and WB is being displaced vertically (upward) ¢sB. Since blocks A and B are
5 A
at rest initially, T1 = 0. Applying Eq. 14–7, we have
T1 + a U1 - 2 = T2

3
1
1
0 + WA ¢ ¢sA ≤ - Ff ¢sA - WB ¢sB = mAv2A + mB v2B
5
2
2
3
1 60
1 10
60 B (3) R - 9.60(3) - 10(6) = ¢
≤ v2A + ¢
≤ v2B
5
2 32.2
2 32.2
1236.48 = 60v2A + 10v2B

(2)

Eqs. (1) and (2) yields
Ans.

vA = 3.52 ft>s
vB = 7.033 ft s

Ans:
vA = 3.52 ft>s
391


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*14–16.
A small box of mass m is given a speed of v = 214gr at the
top of the smooth half cylinder. Determine the angle u at
which the box leaves the cylinder.

A
u

r

O

SOLUTION
Principle of Work and Energy: By referring to the free-body diagram of the block,

Fig. a, notice that N does no work, while W does positive work since it displaces
downward though a distance of h = r - r cos u.
T1 + ©U1 - 2 = T2
1
1
1
m a gr b + mg(r - r cos u) = mv2
2
4
2
v2 = gr a

9
- 2 cos u b
4

2

v
=
Equations of Motion: Here, an =
r
referring to Fig. a,
©Fn = man ;

(1)
gr a

mg cos u - N = m c g a
N = mg a 3 cos u -

9
- 2 cos u b
4
9
= ga - 2 cos u b . By
r
4
9
- 2 cos u b d
4

9
b
4

It is required that the block leave the track. Thus, N = 0.
0 = mg a 3 cos u -

9
b
4

Since mg Z 0,
3 cos u -

9
= 0
4
Ans.

u = 41.41° = 41.4°

Ans:
u = 41.4°
392


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14–17.
If the cord is subjected to a constant force of F = 30 lb and
the smooth 10-lb collar starts from rest at A, determine its
speed when it passes point B. Neglect the size of pulley C.

y ϭ 1 x2
2

y

C

B

4.5 ft
F ϭ 30 lb

A

SOLUTION

x
1 ft

3 ft

2 ft

Free-Body Diagram: The free-body diagram of the collar and cord system at an
arbitrary position is shown in Fig. a.
Principle of Work and Energy: By referring to Fig. a, only N does no work since it
always acts perpendicular to the motion. When the collar moves from position A to
position B, W displaces upward through a distance h = 4.5 ft, while force F displaces a
distance of s = AC - BC = 262 + 4.52 - 2 = 5.5 ft. The work of F is positive,
whereas W does negative work.
TA + gUA - B = TB

0 + 30(5.5) + [- 10(4.5)] =

1 10
a
bv 2
2 32.2 B
Ans.

vB = 27.8 ft>s

Ans:
vB = 27.8 ft>s

393


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14–18.
When the 12-lb block A is released from rest it lifts the two
15-lb weights B and C. Determine the maximum distance
A will fall before its motion is momentarily stopped.
Neglect the weight of the cord and the size of the pulleys.

4 ft

4 ft

Solution
A

Consider the entire system:
t = 2y2 + 42

T1 + ΣU1 - 2 = T2

C

B

(0 + 0 + 0) + 12y - 2(15) ( 2y2 + 42 - 4 ) = (0 + 0 + 0)
0.4y = 2y2 + 16 - 4

(0.4y + 4)2 = y2 + 16

-0.84y2 + 3.20y + 16 = 16
-0.84y + 3.20 = 0
Ans.

y = 3.81 ft

Ans:
y = 3.81 ft
394


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14–19.
If the cord is subjected to a constant force of F = 300 N
and the 15-kg smooth collar starts from rest at A, determine
the velocity of the collar when it reaches point B. Neglect
the size of the pulley.

200 mm
B

C
30Њ
200 mm
F ϭ 300 N

200 mm

SOLUTION

300 mm

Free-Body Diagram: The free-body diagram of the collar and cord system at an
arbitrary position is shown in Fig. a.

A

Principle of Work and Energy: Referring to Fig. a, only N does no work since it
always acts perpendicular to the motion. When the collar moves from position A to
position B, W displaces vertically upward a distance h = (0.3 + 0.2) m = 0.5 m,
while

force
2

F

displaces

a

distance

of

2

s = AC - BC = 20.72 + 0.42 -

20.2 + 0.2 = 0.5234 m . Here, the work of F is positive, whereas W does
negative work.
TA + gUA - B = TB

0 + 300(0.5234) + [-15(9.81)(0.5)] =

1
(15)vB2
2
Ans.

vB = 3.335 m>s = 3.34 m>s

Ans:
vB = 3.34 m>s
395


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*14–20.

Barrier stopping force (kip)

The crash cushion for a highway barrier consists of a nest of

barrels filled with an impact-absorbing material. The barrier
stopping force is measured versus the vehicle penetration
into the barrier. Determine the distance a car having a
weight of 4000 lb will penetrate the barrier if it is originally
traveling at 55 ft>s when it strikes the first barrel.

SOLUTION
T1 + ©U1 - 2 = T2
1 4000
a
b (55)2 - Area = 0
2 32.2
Area = 187.89 kip # ft

36
27
18
9
0

2

5

10
15
20
25
Vehicle penetration (ft)

2(9) + (5 - 2)(18) + x(27) = 187.89
(O.K!)

x = 4.29 ft 6 (15 - 5) ft
Thus

Ans.

s = 5 ft + 4.29 ft = 9.29 ft

Ans:
s = 9.29 ft
396


© 2016 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.

14–21.
Determine the velocity of the 60-lb block A if the two
blocks are released from rest and the 40-lb block B moves
2 ft up the incline. The coefficient of kinetic friction
between both blocks and the inclined planes is mk = 0.10.

SOLUTION
Block A:
+a©Fy = may;

A

NA - 60 cos 60° = 0
NA = 30 lb

B
60

30

FA = 0.1(30) = 3 lb
Block B:
+Q©Fy = may;

NB - 40 cos 30° = 0
NB = 34.64 lb
FB = 0.1(34.64) = 3.464 lb

Use the system of both blocks. NA, NB, T, and R do no work.
T1 + ©U1-2 = T2
(0 + 0) + 60 sin 60°|¢sA| - 40 sin 30°|¢sB| - 3|¢sA| -3.464|¢sB| =

1 60
1 40
a
b v2 + a
b v2
2 32.2 A
2 32.2 B

2sA + sB = l
2¢sA = - ¢sB

When |¢sB| = 2 ft, |¢sA| = 1 ft
Also,
2vA = -vB
Substituting and solving,
Ans.

vA = 0.771 ft>s
vB = -1.54 ft>s

Ans:
vA = 0.771 ft>s
397


© 2016 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.

14–22.
The 25-lb block has an initial speed of v0 = 10 ft>s when it
is midway between springs A and B. After striking spring B,
it rebounds and slides across the horizontal plane toward
spring A, etc. If the coefficient of kinetic friction between
the plane and the block is mk = 0.4, determine the total
distance traveled by the block before it comes to rest.

2 ft
kA = 10 lb/in.

1 ft

kB = 60 lb/in.

v0 = 10 ft/s
A

B

SOLUTION
Principle of Work and Energy: Here, the friction force Ff = mk N = 0.4(25) =
10.0 lb. Since the friction force is always opposite the motion, it does negative work.
When the block strikes spring B and stops momentarily, the spring force does
negative work since it acts in the opposite direction to that of displacement.
Applying Eq. 14–7, we have
Tl + a U1 - 2 = T2

1
1 25
a
b (10)2 - 10(1 + s1) - (60)s21 = 0
2 32.2
2
s1 = 0.8275 ft
Assume the block bounces back and stops without striking spring A. The spring
force does positive work since it acts in the direction of displacement. Applying
Eq. 14–7, we have
T2 + a U2 - 3 = T3
0 +

1
(60)(0.82752) - 10(0.8275 + s2) = 0

2

s2 = 1.227 ft
Since s2 = 1.227 ft 6 2 ft, the block stops before it strikes spring A. Therefore, the
above assumption was correct. Thus, the total distance traveled by the block before
it stops is
Ans.

sTot = 2s1 + s2 + 1 = 2(0.8275) + 1.227 + 1 = 3.88 ft

Ans:
sTot = 3.88 ft
398


© 2016 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.

14–23.
The 8-kg block is moving with an initial speed of 5 m>s. If
the coefficient of kinetic friction between the block and
plane is mk = 0.25, determine the compression in the spring
when the block momentarily stops.

5 m/s
B

2m

kA ϭ 200 N/m

A

Solution
Work. Consider the force equilibrium along y axis by referring to the FBD of the
block, Fig. a
+ c ΣFy = 0;   N - 8(9.81) = 0  N = 78.48 N
Thus, the friction is Ff = mkN = 0.25(78.48) = 19.62 N and Fsp = kx = 200 x.
Here, the spring force Fsp and Ff both do negative work. The weight W and normal
reaction N do no work.
UFsp = -

L0

x

200 x dx = - 100 x2

UFf = - 19.62(x + 2)
Principle of Work And Energy. It is required that the block stopped momentarily,
T2 = 0. Applying Eq. 14–7
T1 + Σ U1 - 2 = T2
1
(8) ( 52 ) +
2

( - 100x2 ) + [ - 19.62(x + 2)] = 0

100x2 + 19.62x - 60.76 = 0
Solved for positive root,

Ans.

x = 0.6875 m = 0.688 m

Ans:
x = 0.688 m
399


© 2016 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.

*14–24.
At a given instant the 10-lb block A is moving downward
with a speed of 6 ft s. Determine its speed 2 s later. Block B
has a weight of 4 lb, and the coefficient of kinetic friction
between it and the horizontal plane is mk = 0.2. Neglect the
mass of the cord and pulleys.

B

SOLUTION

A

Kinematics: The speed of the block A and B can be related by using the position
coordinate equation.
sA + (sA - sB) = l
2¢sA - ¢sB = 0

2sA - sB = l
[1]

¢sB = 2¢sA

[2]

yB = 2yA
Equation of Motion:
+ ©Fy¿ = may¿ ;

NB - 4 =

4
(0)
32.2

NB = 4.00 lb

Principle of Work and Energy: By considering the whole system, WA, which acts
in the direction of the displacement, does positive work. The friction force
Ff = mk NB = 0.2(4.00) = 0.800 lb does negative work since it acts in the opposite
direction to that of displacement. Here, WA is being displaced vertically
(downward) ¢sA. Applying Eq. 14–7, we have
T1 + a U1 - 2 = T2
1
1
m A y2A B 0 + mB A y2B B 0 + WA ¢sA - Ff ¢sB
2 A
2

=

1
1
m y2 + mB y2B
2 A A
2

[3]
(yA)0 + yA
d (2) =
2
(Eq. [2]). Substituting these

From Eq. [1], (yB)0 = 2(yA)0 = 2(6) = 12 ft>s. Also, ¢sA = c
(yA)0 + yA = 6 + yA and ¢sB = 2¢sA = 12 + 2yA
values into Eq. [3] yields

4
1
1 10
a
b A 62 B + a
b A 122 B + 10(6 + yA B - 0.800(12 + 2yA)
2 32.2
2 32.2
=

1 10
1

4
a
b y2A + a
b A 4y2A B
2 32.2
2 32.2
Ans.

yA = 26.8 ft>s

Ans:
vA = 26.8 ft>s
400


© 2016 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.

14–25.
The 5-lb cylinder is falling from A with a speed vA = 10 ft>s
onto the platform. Determine the maximum displacement
of the platform, caused by the collision. The spring has an
unstretched length of 1.75 ft and is originally kept in
compression by the 1-ft long cables attached to the platform.
Neglect the mass of the platform and spring and any energy
lost during the collision.

vA ϭ 10 ft/s

Solution

A

3 ft

T1 + Σ U1 - 2 = T2
1
1
5
1
a
b ( 102 ) + 5(3 + s) - c (400)(0.75 + s)2 - (400)(0.75)2 d = 0
2 32.2
2
2

200 s2 + 295 s - 22.76 = 0

k ϭ 400 lb/ft

(O.K!)

s = 0.0735 ft 6 1 ft

1 ft

Ans.

s = 0.0735 ft

Ans:
s = 0.0735 ft
401