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: 38

Answer

The acceleration of the 2.0-kg block is $2.29~m/s^2$.

Work Step by Step

Note that the three blocks will all have the same magnitude of acceleration and the 3.0-kg will move downward while the 1.0-kg moves upward. We can set up a force equation for the 1.0-kg block. Let $T_1$ be the tension in the rope attached to this block. Let $m_1$ be the mass of this block. $\sum F = m_1~a$ $T_1- m_1~g= m_1~a$ $T_1= m_1~(g+a)$ We can set up a force equation for the 3.0-kg block. Let $T_2$ be the tension in the rope attached to this block. Let $m_3$ be the mass of this block. $\sum F = m_3~a$ $m_3~g - T_2 = m_3~a$ $T_2 = m_3~(g-a)$ To find the acceleration of the 2.0-kg block, we can set up a force equation for the 2.0-kg block. Let $m_2$ be the mass of this block. $\sum F = m_2~a$ $T_2 - T_1 - F_f = m_2~a$ $m_3~(g-a) - m_1~(g+a) - m_2~g~\mu_k = m_2~a$ $m_3~g - m_1~g - m_2~g~\mu_k = m_1~a+m_2~a+m_3~a$ $a = \frac{m_3~g - m_1~g - m_2~g~\mu_k}{ m_1+m_2+m_3}$ $a = \frac{(3.0~kg)(9.80~m/s^2) - (1.0~kg)(9.80~m/s^2) - (2.0~kg)(9.80~m/s^2)(0.30)}{1.0~kg+2.0~kg+3.0~kg}$ $a = 2.29~m/s^2$ The acceleration of the 2.0-kg block is $2.29~m/s^2$
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