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: 23

Answer

The tension in rope 2 is 270 N

Work Step by Step

Let's consider the system of sled A. We can find the acceleration of sled A. $\sum F = m_A~a$ $T_1 - m_A~g~\mu_k = m_A~a$ $a = \frac{T_1 - m_A~g~\mu_k}{m_A}$ $a = \frac{(150~N) - (100~kg)(9.80~m/s^2)(0.10)}{100~kg}$ $a = 0.52~m/s^2$ Since both sleds move together, this will be the acceleration of the system of sled B. Let's consider the system of sled B. We can find the tension $T_2$ in rope 2. $\sum F = m_B~a$ $T_2 - T_1 - m_B~g~\mu_k = m_B~a$ $T_2 = T_1 + m_B~g~\mu_k + m_B~a$ $T_2 = (150~N) + (80~kg)(9.80~m/s^2)(0.10) + (80~kg)(0.52~m/s^2)$ $T_2 = 270~N$ The tension in rope 2 is 270 N.
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