Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 5 - Applications of Integration - 5.4 Work - 5.4 Exercises - Page 387: 7

Answer

$\frac{15}{4}$ $ft-lb$

Work Step by Step

According to Hooke's Law, the force required to maintain a spring stretched $x$ units beyond its natural length (or compressed $x$ units less than its natural length) is proportional to $x$, that is, $f(x)$ = $kx$ Here, the amount stretched is $4$ $in.$ = $\frac{1}{3}$ $ft$ and natural length to $6$ $in.$ = $\frac{1}{2}$ $ft$ beyond its natural length. Therefore $10 = \dfrac{1}{3} k\Rightarrow k=30\Rightarrow f(x)=30x$. The work is $W$ = $\int_a^b f(x)dx$=$\int_0^{1/2}{30x}dx$ = $[15x^{2}]|_0^{1/2}$ = $\frac{15}{4}$ $ft-lb$
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