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 - Conceptual Questions - Page 173: 12



Work Step by Step

Potential energy at position $x_{0}$ = $\frac{1}{2}kx_{0}^{2}$ Potential energy at position $x$ = $\frac{1}{2}kx^{2}$ So, change in potential energy =$\frac{1}{2}k(x^{2}-x_{0}^{2})$. i.e. proportional to $(x^{2}-x_{0}^{2})$.
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