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 130: 20


Please see the work below.

Work Step by Step

We know that according to law of conservation of energy $(\frac{1}{2})mv^2=(\frac{1}{2})Kx^2$ This simplifies to: $x=\sqrt{\frac{m}{K}v}$ We plug in the known values to obtain: $x=(\sqrt{\frac{35000}{2800000}})(7.5)=0.84m$
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