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 140: 83

Answer

a). $5.5m/s^{2}$ b). $172.5N$

Work Step by Step

a). Let a be the acceleration of the system. $F-f_{s}=m_{A}a$ $f_{s}-f_{B}=m_{B}a$ $m_{A}g=N$ $N_{B}=(m_{B}g+N)$ So, $N_{B}=(m_{A}+m_{B})g$ Therefore, $F-f_{B}=(m_{A}+m_{B})a$ $F-coefficient_{k}(m_{A}+m_{B})g=(m_{A}+m_{B})a$ From above, by calculating, we get $a=5.5m/s^{2}$ b). $F-f_{s}=m_{A}a$ $f_{s}=F-m_{A}a=200-(5\times5.5)=172.5N$
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