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: 38


$m = 89.3~kg$

Work Step by Step

We can find the acceleration of the system. $y = \frac{1}{2}at^2$ $a = \frac{2y}{t^2}$ $a = \frac{(2)(1.0~m)}{(6.0~s)^2}$ $a = 0.0556~m/s^2$ Let's consider the system of both blocks. The weight of the block of mass $m$ pulls down on the left while the weight of the 100-kg block pulls down on the right. We can use a force equation to find $m$. $\sum F = (m+100~kg)~a$ $(100~kg)~g-mg = (m+100~kg)~a$ $m = \frac{(100~kg)(g-a)}{g+a}$ $m = \frac{(100~kg)(9.80~m/s^2-0.0556~m/s^2)}{9.80~m/s^2+0.556~m/s^2}$ $m = 89.3~kg$
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