Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 10 - Simple Harmonic Motion and Elasticity - Problems - Page 276: 29


The height the block was dropped is $4.8cm$

Work Step by Step

As the block is dropped, it has a speed of $v_0=0$. We choose the height here $h_0=0$ and since the spring is not strained, $y_0=0$. After the spring is compressed by $y_f=h_f=-2.5cm=-2.5\times10^{-2}m$, the block's speed $v_f=0$ There is no rotational or translational kinetic energy. Since total energy is conserved, $$\frac{1}{2}ky_f^2+mgh_f=0$$ $$h_f=-\frac{ky_f^2}{2mg}$$ We have $k=450N/m$ and the block's weight $mg=0.3\times9.8=2.94N$. Therefore, $$h_f=-0.048m=-4.8cm$$ The height the block was dropped is $h_0-h_f=4.8cm$
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