Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 320: 50


$9$ inches

Work Step by Step

The work required to compress the spring beyond equilibrium can be calculated as: $\text{Work}, W= \int_{m}^{n} kx \ dx $; where $k$ is the spring constant. and $\dfrac{1}{2} k(1.5)^2=18 \implies k =16\ lb/ft$ Now, $F=k x \implies 12 =16 x $ or, $x=0.75 \ ft$ Thus, a weight of $ 12 \ lb$ will stretch the spring by $9$ inches.
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