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: 38



Work Step by Step

As the spring stretches from $y_0=0.12m$ to $y_f=0.23m$, the speed of the sphere changes from $v_0=5.7m/s$ to $v_f=4.8m/s$ and its height changes $h_f-h_0=y_0-y_f=-0.11m$ (because the sphere moves downward). Because energy is conserved, $$\frac{1}{2}mv_0^2+mgh_0+\frac{1}{2}ky_0^2=\frac{1}{2}mv_f^2+mgh_f+\frac{1}{2}ky_f^2$$ $$\frac{1}{2}k(y_f^2-y_0^2)=mg(h_0-h_f)+\frac{1}{2}m(v_0^2-v_f^2)$$ $$k=\frac{mg(h_0-h_f)+\frac{1}{2}m(v_0^2-v_f^2)}{\frac{1}{2}(y_f^2-y_0^2)}$$ $$k=\frac{2mg(h_0-h_f)+m(v_0^2-v_f^2)}{(y_f^2-y_0^2)}$$ We have $m=0.6kg, g=9.8m/s^2, h_0-h_f=0.11m, v_0^2-v_f^2=9.45$ and $y_f^2-y_0^2=0.0385$, so $$k=181N/m$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.