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 138: 59

Answer

a). $97.5N$ b). $82.5N$

Work Step by Step

a). $F_{2}-F_{1}=(m_{1}+m_{2})a$ $F_{2}=F_{1}+(m_{1}+m_{2})a$ $F_{2}=(75+(5+10)\times1.5)N=97.5N$ b). $N_{1}=F_{1}+m_{1}a=[75+(5\times1.5)]N=82.5N$ Also, $N_{2}=F_{2}-m_{2}a=[97.5-(10\times1.5)]N=82.5N$ Hence, the force of contact N between the two blocks is $82.5N$
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