University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 6 - Work and Kinetic Energy - Problems - Exercises - Page 200: 6.88


(a) W = 15.0 J (b) W = 0 (c) W = 10 J

Work Step by Step

(a) $F = \alpha~x~y~\hat{i}$ $W = \int_{0}^{2.00}F~dx$ $W = \int_{0}^{2.00}\alpha~x~y~dx$ $W = \alpha~y\int_{0}^{2.00}x~dx$ $W = (2.50~N/m^2)(3.00~m)\frac{x^2}{2}\vert_{0}^{2.00}$ $W = (2.50~N/m^2)(3.00~m)(2.00~m^2)$ $W = 15.0~J$ (b) Since the force acts at a $90^{\circ}$ angle to the direction of motion, the work done by the force is zero. (c) We can find the work done on the object. $F = \alpha~x~y~\hat{i}$ $W = \int_{0}^{2.00}F~dx$ $W = \int_{0}^{2.00}\alpha~x~y~dx$ $W = \int_{0}^{2.00}\alpha~(x)~(1.5~x)~dx$ $W = (1.5)(2.50~N/m^2)\int_{0}^{2.00}x^2~dx$ $W = (1.5)(2.50~N/m^2)\frac{x^3}{3}\vert_{0}^{2.00}$ $W = (1.5)(2.50~N/m^2)(\frac{8.00}{3}~m^3)$ $W = 10~J$
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