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: 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; $PE_{elastic}=\frac{1}{2}(\frac{2.0\times 10^6(0.5\times 10^{-4})}{3.0\times 10^{-3}})\times 1.0\times 10^{-3}=1.7\times 10^{-2}J$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.