KEY CONCEPT FIGURE 8.15 The restoring force for a spring. Equilibrium length The magnitude of the restoring force is proportional to the displacement of the spring from equilibrium. (а) Equilibrium length (b) (c) The farther the spring C is stretched, the Relaxed spring Fp (N) Relaxed harder it pulls back. spring 2.5 - Spring compressed Ax AX! 2.0 The displacement Ax' and the spring force (Fsp)x are always in opposite directions. sp 1.5 Slope = k = 3.5 N/m 1.0 Ax Ax 0.5 Spring extended TAr (m) 0.8 0.0 0.0 0.2 0.4 0.6 wwwo 4 cm 8 N
KEY CONCEPT FIGURE 8.15 The restoring force for a spring. Equilibrium length The magnitude of the restoring force is proportional to the displacement of the spring from equilibrium. (а) Equilibrium length (b) (c) The farther the spring C is stretched, the Relaxed spring Fp (N) Relaxed harder it pulls back. spring 2.5 - Spring compressed Ax AX! 2.0 The displacement Ax' and the spring force (Fsp)x are always in opposite directions. sp 1.5 Slope = k = 3.5 N/m 1.0 Ax Ax 0.5 Spring extended TAr (m) 0.8 0.0 0.0 0.2 0.4 0.6 wwwo 4 cm 8 N
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Question
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The end of a spring is pulled to the right by 4 cm; the restoring force is
8 N to the left. Given the relationships 8.15c, if the spring is returned to equilibrium and then pushed to the left by 2 cm, the restoring force is
A. 4 N to the left. B. 4 N to the right. C. 8 N to the left.
D. 8 N to the right. E. 16 N to the left. F. 16 N to the right.
Transcribed Image Text:KEY CONCEPT
FIGURE 8.15 The restoring force for a spring.
Equilibrium
length
The magnitude of the restoring
force is proportional to the
displacement of the spring
from equilibrium.
(а)
Equilibrium
length
(b)
(c)
The farther the spring C
is stretched, the
Relaxed spring
Fp (N)
Relaxed
harder it pulls back.
spring
2.5 -
Spring compressed
Ax
AX!
2.0
The displacement Ax'
and the spring force
(Fsp)x are always in
opposite directions.
sp
1.5
Slope = k = 3.5 N/m
1.0
Ax
Ax
0.5
Spring
extended
TAr (m)
0.8
0.0
0.0
0.2
0.4
0.6
wwwo
4 cm
8 N
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