University Physics with Modern Physics (14th Edition)

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

Chapter 9 - Rotation of Rigid Bodies - Problems - Exercises - Page 299: 9.75

Answer

$v = \sqrt{\frac{(2)(m_Bg - m_Ag~\mu_k)~d}{m_A+m_B+\frac{I}{R^2}}}$

Work Step by Step

The kinetic energy of block A, block B, and the pulley will be equal in magnitude to the change in potential energy of block B minus the magnitude of work done by friction. $KE = PE + work$ $\frac{1}{2}m_Av^2+\frac{1}{2}m_Bv^2+\frac{1}{2}I\omega^2 = m_Bgd - m_Ag~\mu_k~d$ $\frac{1}{2}m_Av^2+\frac{1}{2}m_Bv^2+\frac{1}{2}I(\frac{v}{R})^2 = (m_Bg - m_Ag~\mu_k)~d$ $v^2 = \frac{(2)(m_Bg - m_Ag~\mu_k)~d}{m_A+m_B+\frac{I}{R^2}}$ $v = \sqrt{\frac{(2)(m_Bg - m_Ag~\mu_k)~d}{m_A+m_B+\frac{I}{R^2}}}$
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