Answer
$x = 0.46m$
Work Step by Step
$K_{max} = \frac{1}{2}kx^2$
$2K_{max} = kx^2$
$ kx^2 = 2K_{max}$
$ x^2 = \frac{2K_{max}}{k}$
$ x = \sqrt {\frac{2K_{max}}{k}}$
$ x = \sqrt {\frac{2\times 66.9}{640}}$
$ x = 0.457m \approx 0.46m$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 8 - Potential Energy and Conservation of Energy - Problems - Page 206: 53c
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