Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 7 - Exercises and Problems - Page 132: 53

Answer

a) 20.1 meters per second b) 30.4 meters per second

Work Step by Step

We know that the energy lost to friction is: a) $E=F_n\mu \Delta x = mgcos\theta \mu \frac{25}{sin\theta}=\frac{25mg\mu}{tan\theta}$ Thus, we find: $\frac{1}{2}mv^2 = -\frac{25mg\mu}{tan\theta}+mgh$ $v= \sqrt{2(-\frac{25g\mu}{tan\theta}+gh)}$ $v=20.1\ m/s$ b) We use a similar process: $E=F_n\mu \Delta x = mgcos\theta \mu \frac{38}{sin\theta}=\frac{38mg\mu}{tan\theta}$ Thus, we find: $\frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 = -\frac{38mg\mu}{tan\theta}+mgh$ $v= \sqrt{2(-\frac{38g\mu}{tan\theta}+gh)+v_0^2}$ $v = 30.4 \ m/s$
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