A force of 7 lb 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? Step 1 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 3/4 ft, we have 7 = So 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. W = dx

A force of 7 lb 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? Step 1 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 3/4 ft, we have 7 = So 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. W = dx

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

6th Edition

ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

ChapterA: Appendix

SectionA.2: Geometric Constructions

Problem 10P: A soda can has a volume of 25 cubic inches. Let x denote its radius and h its height, both in...

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