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: 55

Answer

The acceleration of the 3.0-kg block is $\frac{2m_2~g}{m_1+4m_2}$.

Work Step by Step

We can set up a force equation for the 1.0-kg block. Let $m_2$ be the mass of this block. Let $T$ be the tension in the rope. So; $\sum F = m_2~a_2$ $m_2~g - T = m_2~a_2$ $T = m_2~g - m_2~a_2$ We can use this expression for the tension $T$ in the force equation for 3.0-kg block. Let $m_1$ be the mass of this block. Note that $a_2 = 2a_1$. Then; $\sum F = m_1~a_1$ $2T = m_1~a_1$ $2m_2~g - 2m_2~a_2 = m_1~a_1$ $2m_2~g - 2m_2~(2~a_1) = m_1~a_1$ $2m_2~g = m_1~a_1+4m_2~a_1$ $a_1 = \frac{2m_2~g}{m_1+4m_2}$ The acceleration of the 3.0-kg block is $\frac{2m_2~g}{m_1+4m_2}$.
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