College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 5 - Work and Energy - Learning Path Questions and Exercises - Exercises - Page 176: 28


$E_{1}=10J$ $E_{2}=20J$

Work Step by Step

$x_{1}=\frac{F}{k_{1}}=\frac{100}{500}=0.2m$ $x_{2}=\frac{F}{k_{2}}=\frac{100}{250}=0.4m$ Therefore, $E_{1}=\frac{k_{1}x_{1}^{2}}{2}=\frac{500\times 0.2^{2}}{2}=10J$ $E_{2}=\frac{k_{2}x_{2}^{2}}{2}=\frac{250\times 0.4^{2}}{2}=20J$
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