Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 7 - Exercises and Problems - Page 130: 14

Answer

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: $U=\frac{1}{2}(320)(0.18)^2$ $U=5.2J$
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