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: 55

Answer

a). $1.815m/s^{2}$ b). $6.38N$

Work Step by Step

a). $T-m_{1}g=m_{1}a$ $m_{2}g-T=m_{2}a$ From above 2 equations, $a=\frac{m_{2}-m_{1}}{m_{1}+m_{2}}g$ $a=\frac{0.80-0.55}{0.80+0.55}9.8=1.815m/s^{2}$ b). So, $T=m_{1}(a+g)=0.55(1.815+9.8)=6.38N$
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