## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 11 - Work - Exercises and Problems: 12

#### Answer

The work done by $T_1$ is 1690 J The work done by $T_2$ is 1065 J The work done by $T_3$ is -1950 J

#### Work Step by Step

We can find the work done by $T_1$. $W = T_1\cdot d$ $W = T_1 ~d~cos(\theta)$ $W = (600~N)(3.0~m)~cos(20^{\circ})$ $W = 1690~J$ The work done by $T_1$ is 1690 J. We can find the work done by $T_2$. $W = T_2\cdot d$ $W = T_2 ~d~cos(\theta)$ $W = (410~N)(3.0~m)~cos(30^{\circ})$ $W = 1065~J$ The work done by $T_2$ is 1065 J. We can find the work done by $T_3$. $W = T_3\cdot d$ $W = T_3 ~d~cos(\theta)$ $W = (650~N)(3.0~m)~cos(180^{\circ})$ $W = -1950~J$ The work done by $T_3$ is -1950 J.

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