University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 6 - Work and Kinetic Energy - Problems - Exercises - Page 195: 6.23

Answer

The minimum speed of the box at the bottom of the incline is $v = \sqrt{2gh\times (1 + \mu_k~cot(\alpha))}$.

Work Step by Step

$KE_1 + W = KE_2$ $\frac{1}{2}mv^2 - mgh - F_f~d = 0$ $\frac{1}{2}mv^2 = mgh + mg~\mu_k~cos(\alpha)~\frac{h}{sin(\alpha)}$ $v^2 = 2gh + 2g~\mu_k~cot(\alpha)~h$ $v = \sqrt{2gh\times (1 + \mu_k~cot(\alpha))}$ The minimum speed of the box at the bottom of the incline is $v = \sqrt{2gh\times (1 + \mu_k~cot(\alpha))}$.
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