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 5 - Force and Motion - Exercises and Problems: 9

Answer

$\frac{m_1}{m_2} = 0.48$

Work Step by Step

Let $F$ be the force provided by one rubber band. From graph 1, we can see that the acceleration of $m_1$ is $5a$ when there are three rubber bands. We can set up an equation. $3F = (m_1)(5a)$ $m_1 = \frac{3F}{5a}$ From graph 2, we can see that the acceleration of $m_2$ is $4a$ when there are five rubber bands. We can set up an equation. $5F = (m_2)(4a)$ $m_2 = \frac{5F}{4a}$ We can find the mass ratio $\frac{m_1}{m_2}$ as: $\frac{m_1}{m_2} = \frac{3F/5a}{5F/4a}$ $\frac{m_1}{m_2} = \frac{12Fa}{25Fa}$ $\frac{m_1}{m_2} = \frac{12}{25}$ $\frac{m_1}{m_2} = 0.48$
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