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: 6.16

Answer

(a) The work done was -6.56 J. (b) v = 0.750 m/s (c) v = 1.60 m/s

Work Step by Step

(a) We can find the kinetic energy at point A. $KE_A = \frac{1}{2}mv^2 = \frac{1}{2}(1.50~kg)(3.21~m/s)^2$ $KE_A = 7.728~J$ We can find the kinetic energy at point B. $KE_B = \frac{1}{2}mv^2 = \frac{1}{2}(1.50~kg)(1.25~m/s)^2$ $KE_B = 1.172~J$ We can find the change in kinetic energy. $\Delta KE = KE_B-KE_A = 1.172~J - 7.728~J$ $\Delta KE = -6.56~J$ The work done was -6.56 J. (b) We can find the kinetic energy at point C. $KE_C - KE_B = -0.750~J$ $KE_C = KE_B-0.750~J$ $KE_C = 1.172~J-0.750~J$ $KE_C = 0.422~J$ We can find the speed $v$ at point C. $ \frac{1}{2}mv^2 = 0.422~J$ $v^2 = \frac{0.844~J}{1.50~kg}$ $v = \sqrt{\frac{0.844~J}{1.50~kg}}$ $v = 0.750~m/s$ (c) We can find the kinetic energy at point C. $KE_C - KE_B = 0.750~J$ $KE_C = KE_B+0.750~J$ $KE_C = 1.172~J+0.750~J$ $KE_C = 1.922~J$ We can find the speed $v$ at point C. $ \frac{1}{2}mv^2 = 1.922~J$ $v^2 = \frac{3.844~J}{1.50~kg}$ $v = \sqrt{\frac{0.844~J}{1.50~kg}}$ $v = 1.60~m/s$
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