Answer
$m=1.39Kg$
Work Step by Step
As $K.E_{max}=E_{mec}$
but $K.E_{max}=\frac{1}{2}mv_{max}^2$
Hence
$\frac{1}{2}mv_{max}^2=E_{mec}$
$m=\frac{2E_{mec}}{v_{max}^2}$
putting the values, we get
$m=\frac{2\times1}{(1.2)^2}$
$m=1.39Kg$
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: 30b
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