## College Physics (7th Edition)

Published by Pearson

# Chapter 4 - Force and Motion - Learning Path Questions and Exercises - Exercises - Page 138: 61

#### Answer

a). $1.238m/s^{2}$ b). $21.4N$

#### Work Step by Step

$m_{1}=3kg$ $m_{2}=2.5kg$ a). $T-m_{1}gsin37^{\circ}=m_{1}a$ $T=m_{1}gsin37^{\circ}+m_{1}a$ , equation (1) Also, $m_{2}g-T=m_{2}a$ , equation (2) From equation (1) and (2), $a=\frac{m_{2}g-m_{1}gsin37^{\circ}}{m_{1}+m_{2}}=1.238m/s^{2}$ b). $T=m_{2}(g-a)=2.5\times(9.8-1.238)=21.4N$

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