Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 7 - Exercises and Problems: 64

Answer

Please see the work below.

Work Step by Step

According to law of conservation of energy $\frac{1}{2}mv^2=\frac{1}{2}kx_{\circ}^2$ This simplifies to: $k=\frac{mv^2}{x_{\circ}^2}$ We plug in the known values to obtain: $k=\frac{1500(2400)^2}{(17)^2}=29.9M\frac{N}{m}$
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