Answer
Please see the work below.
Work Step by Step
We know that
$\frac{1}{2}Kx^2=mgdsin(\theta)$
This simplifies to:
$d=\frac{Kx^2}{2\space m\space g\space sin(\theta)}$
Essential University Physics: Volume 1 (4th Edition)
by Wolfson, Richard
Published by
Pearson
ISBN 10:
0-134-98855-8
ISBN 13:
978-0-13498-855-9
Chapter 7 - Exercises and Problems - Page 131: 41
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