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PROBLEM 13.68
A spring is used to stop a 50-kg package which is moving down a
20º incline. The spring has a constant k = 30 kN/m and is held by
cables so that it is initially compressed 50 mm. Knowing that the
velocity of the package is 2 m/s when it is 8 m from the spring and
neglecting friction, determine the maximum additional deformation
of the spring in bringing the package to rest.
SOLUTION
Let position 1 be the starting position 8 m from the end of the spring when it is compressed 50 mm by the
cable. Let position 2 be the position of maximum compression. Let x be the additional compression of the
spring. Use the principle of conservation of energy. T1 + V1 = T2 + V2 .
Position 1:
1 2 1
mv1 = (50)(2)2 = 100 J
2
2
V1g = mgh1 = (50)(9.81)(8 sin 20°) = 1342.09 J
T1 =
V1e =
1 2 1
ke1 = (30 × 103 )(0.050) 2 = 37.5 J
2
2
1 2
mv2 = 0 since v2 = 0.
2
= mgh2 = (50)(9.81)(− x sin 20°) = −167.76 x
T2 =
Position 2:
V2 g
V2e =
1 2 1
ke2 = (30 × 103 )(0.05 + x)2 = 37.5 + 1500 x + 15000 x 2
2
2
Principle of conservation of energy:
100 + 1342.09 + 37.5 = −167.61x + 37.5 + 1500 x + 15000 x 2
15, 000 x 2 + 1332.24 x − 1442.09 = 0
Solving for x,
x = 0.26882 and − 0.357 64
x = 0.269 m
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605
PROBLEM 13.69
Solve Problem 13.68 assuming the kinetic coefficient of friction between
the package and the incline is 0.2.
PROBLEM 13.68 A spring is used to stop a 50-kg package which is
moving down a 20° incline. The spring has a constant k = 30 kN/m
and is held by cables so that it is initially compressed 50 mm. Knowing
that the velocity of the package is 2 m/s when it is 8 m from the spring
and neglecting friction, determine the maximum additional deformation
of the spring in bringing the package to rest.
SOLUTION
Let position 1 be the starting position 8 m from the end of the spring when it is compressed 50 mm by the
cable. Let position 2 be the position of maximum compression. Let x be the additional compression of the
spring. Use the principle of work and energy. T1 + V1 + U1→2 = T2 + V2
Position 1.
1 2 1
mv1 = (50)(2) 2 = 100 J
2
2
V1g = mgh1 = (50)(9.81)(8sin 20°) = 1342.09 J
T1 =
V1e =
Position 2.
1 2 1
ke1 = (30 × 103 )(0.05) 2 = 37.5 J
2
2
1 2
mv2 = 0 since v2 = 0.
2
= mgh2 = (50)(9.81)(− x sin 20°) = −167.76 x
T2 =
V2 g
V2e =
1 2 1
ke2 = (30 × 103 )(0.05 + x) 2 = 37.5 + 1500 x + 15,000 x 2
2
2
Work of the friction force.
ΣFn = 0
N − mg cos 20° = 0
N = mg cos 20°
= (50)(9.81) cos 20°
= 460.92 N
F f = µk N
U1→2
= (0.2)(460.92)
= 92.184
= − Ff d
= −92.184(8 + x)
= −737.47 − 92.184 x
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reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
606
PROBLEM 13.69 (Continued)
Principle of work and energy:
T1 + V1 + U1− 2 = T2 + V2
100 + 1342.09 + 37.5 − 737.47 − 92.184 x
= −167.76 x + 37.5 + 1500 x + 15, 000 x 2
15, 000 x 2 + 1424.42 x − 704.62 = 0
Solving for x,
x = 0.17440 and −0.26936
x = 0.1744 m
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
607
PROBLEM 13.70
A section of track for a roller coaster consists of two circular
arcs AB and CD joined by a straight portion BC. The radius of
AB is 27 m and the radius of CD is 72 m. The car and its
occupants, of total mass 250 kg, reach Point A with practically
no velocity and then drop freely along the track. Determine the
normal force exerted by the track on the car as the car reaches
point B. Ignore air resistance and rolling resistance.
SOLUTION
Calculate the speed of the car as it reaches Point B using the principle of conservation of energy as the car
travels from position A to position B.
Position A:
v A = 0,
TA =
Position B:
VB = −mgh
1 2
mv A = 0, VA = 0 (datum)
2
where h is the decrease in elevation between A and B.
TB =
Conservation of energy:
1 2
mvB
2
TA + VA = TB + VB :
1 2
mvB − mgh
2
vB2 = 2 gh
0+0=
= (2)(9.81 m/s 2 )(27 m)(1 − cos 40°)
= 123.94 m 2 /s 2
Normal acceleration at B:
( aB ) n =
vB2
ρ
=
123.94 m 2 /s 2
= 4.59 m/s 2
27 m
(a B )n = 4.59 m/s 2
50°
Apply Newton’s second law to the car at B.
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
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