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 - Exercises and Problems - Page 131: 38


Please see the work below.

Work Step by Step

We know that (a) $K=\frac{F}{x}$ $K=\frac{mg}{x}$ $K=\frac{(125)(9.8)}{0.00266}=4.6\times 10^5\frac{N}{m}$ (b) We know that $W=\frac{1}{2}kx^2$ This can be rearranged as: $x=\sqrt{\frac{2W}{x}}$ We plug in the known values to obtain: $x=\sqrt{\frac{(2)(50.0)}{4.6\times 10^5}}$ $x=0.015m$
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