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

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 9 - Work and Kinetic Energy - Exercises and Problems: 34

Answer

(a) $W_T = 570~J$ $W_g = -200~J$ The normal force does zero work on the crate. (b) The increase in thermal energy is 85 J

Work Step by Step

(a) We can find the work done by tension; $W_T = T\cdot d$ $W_T = T~d~cos(\theta)$ $W_T = (120~N)(5.0~m)~cos(18^{\circ})$ $W_T = 570~J$ We can find the work done by gravity. $W_g = F_g \cdot d$ $W_g = (mg)~d~cos(\theta)$ $W_g = (8.0~kg)(9.80~m/s^2)(5.0~m)~cos(120^{\circ})$ $W_g = -200~J$ The normal force acts at a $90^{\circ}$ angle to the direction of motion. Therefore, the normal force does zero work on the crate. (b) The increase in thermal energy is equal in magnitude to the work done by friction; $W_f = F_f\cdot d$ $W_f = (mg)~cos(\theta)~\mu_k~d~cos(180^{\circ})$ $W_f = (8.0~kg)(9.80~m/s^2)~cos(30^{\circ})(0.25)(5.0~m)(-1)$ $W_f = -85~J$ The increase in thermal energy is 85 J
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