A force of 7 Ib is required to hold a spring stretched 9 in. beyond its natural length. How much work W is done in stretching it from its natural length to 15 in. beyond its natural length?

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
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Author:Bruce Crauder, Benny Evans, Alan Noell
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ChapterA: Appendix
SectionA.2: Geometric Constructions
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A force of 7 Ib is required to hold a spring stretched 9 in. beyond its natural length. How much work W is done
in stretching it from its natural length to 15 in. beyond its natural length?
According to Hooke's Law, the force required to maintain a spring stretched x units beyond its natural length is
proportional to x. This means that f(x) = kx, for some constant k.
Since our spring is stretched 9 in., which is equal to 3/4
3
2k. So
4
3/4 ft, we have 7 =
k = 28/3
28/3 Ib/ft.
Now, we can say that the work done in stretching the spring 15 in. =
ft beyond its natural
length is given by the following.
15/12
28
X-
3
W =
dx
Transcribed Image Text:A force of 7 Ib is required to hold a spring stretched 9 in. beyond its natural length. How much work W is done in stretching it from its natural length to 15 in. beyond its natural length? According to Hooke's Law, the force required to maintain a spring stretched x units beyond its natural length is proportional to x. This means that f(x) = kx, for some constant k. Since our spring is stretched 9 in., which is equal to 3/4 3 2k. So 4 3/4 ft, we have 7 = k = 28/3 28/3 Ib/ft. Now, we can say that the work done in stretching the spring 15 in. = ft beyond its natural length is given by the following. 15/12 28 X- 3 W = dx
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