## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 9 - Work and Kinetic Energy - Exercises and Problems - Page 228: 19

#### Answer

The work done by $T_1$ is -7920 J The work done by $T_2$ is -4580 J The work done by gravity is 12,500 J

#### Work Step by Step

We can find the work done by $T_1$ as; $W = T_1\cdot d$ $W = T_1 ~d~cos(\theta)$ $W = (1830~N)(5.00~m)~cos(150^{\circ})$ $W = -7920~J$ The work done by $T_1$ is -7920 J We can find the work done by $T_2$ as; $W = T_2\cdot d$ $W = T_2 ~d~cos(\theta)$ $W = (1295~N)(5.00~m)~cos(135^{\circ})$ $W = -4580~J$ The work done by $T_2$ is -4580 J We can find the work done by gravity as; $W_g = F\cdot d$ $W_g = F ~d~cos(\theta)$ $W_g = (mg) ~d~cos(0^{\circ})$ $W_g = (255~kg)(9.80~m/s^2)(5.00~m)(1)$ $W_g = 12,500~J$ The work done by gravity is 12,500 J

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