Answer
$P.E_i=125J$
Work Step by Step
The initial potential energy $'P.E_i'$ is given as
$P.E_i=\frac{1}{2}KX^2$
given that $X=50cm=\frac{50}{100}m=0.5m$
putting the values, we get
$P.E_i=\frac{1}{2}(1000)(0.5)^2$
$P.E_i=125J$
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 15 - Oscillations - Problems - Page 437: 31b
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