Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 10 - Interactions and Potential Energy - Exercises and Problems - Page 258: 46


The spring constant of the spring is $1.96\times 10^5~N/m$

Work Step by Step

The energy stored in the spring when it is compressed will be equal to the initial potential energy of the safe. Note that the initial height of the safe is 2.5 meters above its final position. Therefore; $U_s = PE$ $\frac{1}{2}kx^2 = mgh$ $k = \frac{2mgh}{x^2}$ $k = \frac{(2)(1000~kg)(9.80~m/s^2)(2.5~m)}{(0.50~m)^2}$ $k = 1.96\times 10^5~N/m$ The spring constant of the spring is $1.96\times 10^5~N/m$
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