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: 32



Work Step by Step

Before the trigger is pulled, we choose the height $h_0=0$ and the spring is compressed by $y_0=9.1\times10^{-2}m$. When the pellet rises to its maximum height $h_f=6.1m$, the spring is completely released and unstrained, so $y_f=0$ At both points, the pellet has zero speed, so $KE=0$. There is also no rotational kinetic energy. Since total energy is conserved, $$\frac{1}{2}ky_0^2=mgh_f$$ $$k=\frac{2mgh_f}{y_0^2}$$ The pellet's weight $mg=(2.1\times10^{-2}kg)\times(9.8m/s^2)=0.206N$. Therefore, $$k=303N/m$$
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