Physics (10th Edition)

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

Chapter 6 - Work and Energy - Problems - Page 166: 20



Work Step by Step

From the given information, $v_0=0$, $v_f=2m/s$ and $m_{sled}=16kg$. Using these, we can calculate the kinetic energies: $KE_0=\frac{1}{2}mv_0^2=0$ $KE_f=\frac{1}{2}mv_f^2=32J$ According to the work-energy theorem, $W=KE_f-KE_0=32J$ We have $$W=(\sum F\cos\theta)s=(F-f_k)\cos\theta\times s$$ We know $s=8m$ and $\sum F$ is parallel with displacement, so $\theta=0$. Therefore, $$F-f_k=\frac{W}{s}=4N$$ $$f_k=F-4=24-4=20N$$ We have $f_k=\mu_kF_N$. Since there is no vertical acceleration, $F_N=mg=16\times9.8=156.8N$ $$156.8\mu_k=20$$ $$\mu_k=0.128$$
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.