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 195: 6.14


W = 135 J

Work Step by Step

We can find the magnitude of the force. $F = \sqrt{(-68.0~N)^2+(36.0~N)^2}$ $F = 76.9~N$ We can find the angle $\alpha$ above the (-x)-axis. $tan(\alpha) = \frac{36.0~N}{68.0~N}$ $\alpha = arctan(\frac{36.0~N}{68.0~N})$ $\alpha = 27.9^{\circ}$ We can find the angle $\theta$ between the force vector and the displacement vector. $\theta = 27.9^{\circ}+60.0^{\circ}$ $\theta = 87.9^{\circ}$ We can find the work done by the force. $W = F~d~cos(\theta)$ $W = (76.9~N)(48.0~m)~cos(87.9^{\circ})$ $W = 135~J$
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