Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 6 - Work and Energy - Problems - Page 167: 59

Answer

(a) The original kinetic energy of the player is 1060 J. (b) The average power required to stop the player is 1060 W.

Work Step by Step

(a) $KE = \frac{1}{2}mv^2$ $KE = \frac{1}{2}(85 ~kg)(5.0 ~m/s)^2$ $KE = 1060 ~J$ The original kinetic energy of the player is 1060 J. (b) $P = \frac{KE}{t}$ $P = \frac{1060 ~J}{1 ~s}$ $P = 1060 ~W$ The average power required to stop the player is 1060 W.
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