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: 33


(a) $T = 3.92~N$ (b) $a = 2.16~m/s^2$

Work Step by Step

(a) Since the 1.0-kg block does not move, the tension $T$ in the rope attached to the 1.0-kg block must be equal in magnitude to the force of friction exerted on the 1.0-kg block. $T = F_f$ $T = (1.0~kg)(9.80~m/s^2)(0.40)$ $T = 3.92~N$ (b) Let's consider the system of the 2.0-kg block. We can use a force equation to find the acceleration of the 2.0-kg block. $\sum F = (2.0~kg)~a$ $F - F_{f1}-F_{f2}= (2.0~kg)~a$ $a = \frac{F - F_{f1}-F_{f2}}{2.0~kg}$ $a = \frac{20~N - (1.0~kg)(9.80~m/s^2)(0.40)-(3.0~kg)(9.80~m/s^2)(0.40)}{2.0~kg}$ $a = 2.16~m/s^2$
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