Essential University Physics: Volume 1 (4th Edition)

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

Chapter 7 - Section 7.3 - Conservation of Mechanical Energy - Example - Page 121: 7.5


1.7 meters

Work Step by Step

We know from the situation that: $mgh = \frac{1}{2} kx^2$ Thus, we find: $ h = \frac{kx^2}{2mg}$ $ h = \frac{140(.11)^2}{2(.05)(9.81)}=1.7m$
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