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 198: 6.63

Answer

(a) W 608 J (b) W = -395 J (c) W = 0 (d) W = -189 J (e) W = 24 J (f) v = 1.55 m/s

Work Step by Step

(a) $W = F~d = (160~N)(3.80~m)$ $W = 608~J$ (b) $W = -mg~sin(32.0^{\circ})~d$ $W = -(20.0~kg)(9.80~m/s^2)~sin(32.0^{\circ})(3.80~m)$ $W = -395~J$ (c) Since the normal force acts at an angle of $90^{\circ}$ to the direction of motion, the normal force does zero work. (d) $W = -F_f~d$ $W = -mg~cos(\theta)~\mu_k~d$ $W = -(20.0~kg)(9.80~m/s^2)~cos(32.0^{\circ})(0.300)(3.80~m)$ $W = -189~J$ (e) $W_{tot} = 608~J -395~J-189~J$ $W_{tot} = 24~J$ (f) $\frac{1}{2}mv^2 = 24~J$ $v^2 = \frac{48~J}{m}$ $v = \sqrt{\frac{48~J}{20.0~kg}}$ $v = 1.55~m/s$
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