Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 7 - Exercises and Problems - Page 125: 16


Please see the work below.

Work Step by Step

We can find the potential energy as follows: $U=\frac{1}{2}Kx^2$ We plug in the known values to obtain: $U=\frac{1}{2}(320)(0.18)^2$ $U=5.2J$
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.