Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 7 - Newton's Third Law - Exercises and Problems - Page 188: 35


$a = 1.77~m/s^2$

Work Step by Step

Note that the 1.0-kg block has the same magnitude of acceleration as the 2.0-kg block. We can set up a force equation for the 1.0-kg block. Let $m_1$ be the mass of this block. $\sum F = m_1~a$ $T-F_f = m_1~a$ $T-m_1~g~\mu_k = m_1~a$ $T = m_1~g~\mu_k + m_1~a$ We can use the expression for tension $T$ in the force equation for the 2.0-kg block. Let $m_2$ be the mass of this block. $\sum F = m_2~a$ $F - T - F_{f1}- F_{f2} = m_2~a$ $F - (m_1~g~\mu_k + m_1~a) - m_1~g~\mu_k- (m_1+m_2)~g~\mu_k = m_2~a$ $F - 3m_1~g~\mu_k - m_2~g~\mu_k = (m_1+m_2)~a$ $a = \frac{F - 3m_1~g~\mu_k - m_2~g~\mu_k}{m_1+m_2}$ $a = \frac{20~N - (3)(1.0~kg)(9.80~m/s^2)(0.30) - (2.0~kg)(9.80~m/s^2)(0.30)}{1.0~kg+2.0~kg}$ $a = 1.77~m/s^2$
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