Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 5 - Exercises and Problems - Page 88: 66


The proof is below.

Work Step by Step

We know that the force of friction is given by: $\mu_k F_n=mu_k mgcos\theta$ We use conservation of energy to find: $mgh=\frac{1}{2}mv^2 - F_n \mu \Delta x$ $mgh=\frac{1}{2}mv^2 - mg \mu h\frac{1}{tan\theta}$ We know that at the bottom of the ramp: $\frac{1}{2}m(\frac{1}{2}v)^2=mgh - mg \mu h\frac{1}{tan\theta}$ Using substitution, it follows: $\frac{1}{2}m(\frac{1}{2}v)^2=\frac{1}{2}mv^2 - mg\mu h\frac{1}{tan\theta} - mg\mu h\frac{1}{tan\theta}$ $-\frac{3}{8}v^2 =- 2g \mu h\frac{1}{tan\theta} $ We need to find a value for h. Thus, we obatin: $mgh =\frac{1}{2}mv^2 - mg \mu h\frac{1}{tan\theta} $ $h =\frac{1}{2g}v^2 - \mu h\frac{1}{tan\theta}$ $h + \mu h\frac{1}{tan\theta}=\frac{1}{2g}v^2 $ $h (1+ \mu \frac{1}{tan\theta})=\frac{v^2}{2g} $ $h = \frac{v^2}{2g(1+ \mu \frac{1}{tan\theta})}$ Using this, we find: $-\frac{3}{8}v^2 =- 2g \mu (\frac{v^2}{2g(1+ \mu \frac{1}{tan\theta})})\frac{1}{tan\theta} $ $-\frac{3}{8}=\frac{-\mu \frac{1}{tan\theta}}{1+ \mu \frac{1}{tan\theta}}$ $(1+ \mu \frac{1}{tan\theta})(-\frac{3}{8})=-\mu \frac{1}{tan\theta}$ $\frac{-3}{8}=-\frac{5}{8} \mu \frac{1}{tan\theta}$ $\mu=\frac{3}{5}tan\theta$ Hence, the proof is completed.
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