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 5 - Force and Motion - Exercises and Problems - Page 127: 7


(a) The acceleration is $2.4~m/s^2$ (b) The acceleration is $0.60~m/s^2$

Work Step by Step

Let $F$ be the force provided by one rubber band. With the information in the question, we can set up a force equation; $2F = Ma$ $F = \frac{M~(1.2~m/s^2)}{2} = M~(0.60~m/s^2)$ (a) $4F = 4\times M~(0.60~m/s^2) = M~(2.4~m/s^2)$ The acceleration is $2.4~m/s^2$ (b) $2F = 2\times M~(0.60~m/s^2) = (2M)(0.60~m/s^2)$ The acceleration is $0.60~m/s^2$
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