Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 7 - Exercises and Problems - Page 131: 41

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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