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 199: 6.74

Answer

(a) The total work done is -910 J. (b) The work done by the cyclist is 3170 J.

Work Step by Step

(a) $W_{tot} = K_2-K_1$ $W_{tot} = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2$ $W_{tot} = \frac{1}{2}(80.0~kg)(1.50~m/s)^2-\frac{1}{2}(80.0~kg)(5.00~m/s)^2$ $W_{tot} = -910~J$ (b) Let $W_C$ be the work done by the cyclist. $W_{tot} = W_g+W_C$ $W_C = W_{tot}-W_g$ $W_C = -910~J - (-(80.0~kg)(9.80~m/s^2)(5.20~m))$ $W_C = 3170~J$ The work done by the cyclist is 3170 J.
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