Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 9 - Static Equilibrium; Elasticity and Fracture - Problems - Page 256: 49

Answer

Please see the work below.

Work Step by Step

We know that $PE_{elastic}=\frac{1}{2}(\frac{EA}{L_{\circ}}dL)$ Thus; $PE_{elastic}=\frac{1}{2}(\frac{2.0\times 10^6(0.5\times 10^{-4})}{3.0\times 10^{-3}})\times 1.0\times 10^{-3}=1.7\times 10^{-2}J$
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