College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 4 - Force and Motion - Learning Path Questions and Exercises - Exercises - Page 137: 46


a). $2.45m/s^{2}$, b). $-1.96m/s^{2}$

Work Step by Step

Let tension in rope between $m_{1}$ and $m_{3}$ be $T_{1}$. Let tension in rope between $m_{3}$ and $m_{2}$ be $T_{2}$. $T_{1}-m_{1}g=m_{1}a$ $T_{1}=m_{1}(a+g)$ $m_{2}g-T_{2}=m_{2}a$ $T_{2}=m_{2}(g-a)$ $T_{2}-T_{1}=m_{3}a$ So, $m_{2}(g-a) -m_{1}(a+g)=m_{3}a$ $a(m_{1}+m_{2}+m_{3})=(m_{2}-m_{1})g$ or, $a=\frac{(m_{2}-m_{1})g}{(m_{1}+m_{2}+m_{3})}$ a). $m_{1}=0.25kg$, $m_{2}=0.5kg$, $m_{3}=0.25kg$, $a=2.45m/s^{2}$ b). $m_{1}=0.35kg$, $m_{2}=0.15kg$, $m_{3}=0.5kg$, $a=-1.96m/s^{2}$, in this case $m_{3}$ will accelerate towards the left.
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