Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 8 - Potential Energy and Conservation of Energy - Problems - Page 212: 130l

Answer

$\Delta U_e = 4.0~J$

Work Step by Step

We can find the change in elastic potential energy: $\Delta U_e = \frac{1}{2}ky_2^2-\frac{1}{2}ky_1^2$ $\Delta U_e = \frac{1}{2}ky_2^2-0$ $\Delta U_e = \frac{1}{2}(200~N/m)(-0.20~m)^2$ $\Delta U_e = 4.0~J$
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.