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: 39

Answer

Please see the work below.

Work Step by Step

We know that $\frac{1}{2}Kx^2=mgdsin(\theta)$ This simplifies to: $d=\frac{Kx^2}{2\space m\space g\space sin(\theta)}$
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