Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 7 - Newton's Third Law - Exercises and Problems: 54

Answer

The acceleration of $m_1$ is $\frac{2m_2~g}{4m_1 + m_2}$.

Work Step by Step

We can set up a force equation for $m_2$. Let $T$ be the tension in the rope. So; $\sum F = m_2~a_2$ $m_2~g - 2T = m_2~a_2$ $T = \frac{m_2~g - m_2~a_2}{2}$ We can use this expression for the tension $T$ in the force equation for $m_1$. Note that $a_2 = \frac{a_1}{2}$. Then; $\sum F = m_1~a_1$ $T = m_1~a_1$ $\frac{m_2~g - m_2~a_2}{2} = m_1~a_1$ $m_2~g - m_2~(\frac{a_1}{2}) = 2m_1~a_1$ $2m_2~g - m_2~a_1 = 4m_1~a_1$ $2m_2~g = 4m_1~a_1 + m_2~a_1$ $a_1 = \frac{2m_2~g}{4m_1 + m_2}$ The acceleration of $m_1$ is $\frac{2m_2~g}{4m_1 + m_2}$.
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