College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Review & Synthesis: Chapters 6-8 - Review Exercises - Page 324: 23


The block's speed is $~1.99~m/s$

Work Step by Step

We can use energy and work to find the speed of the block: $KE_f = U_0+Work$ $\frac{1}{2}mv^2 = mgh-mg~cos~\theta~\mu_k~d$ $\frac{1}{2}mv^2 = mgd~sin~\theta-mg~\mu_k~d~cos~\theta$ $v^2 = 2gd~(sin~\theta-\mu_k~cos~\theta)$ $v = \sqrt{2gd~(sin~\theta-\mu_k~cos~\theta)}$ $v = \sqrt{(2)(9.80~m/s^2)(0.30~m)~[sin~60.0^{\circ}-(0.38)~cos~60.0^{\circ}]}$ $v = 1.99~m/s$ The block's speed is $~1.99~m/s$.
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