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 - General Problems - Page 167: 73

Answer

The final speed of the crate is 14.9 m/s.

Work Step by Step

We can use work and energy to solve this question. Note that the initial kinetic energy is zero. The final kinetic energy is the sum of the positive work done by the pull $W_p$ and the negative work of friction $W_f$. Therefore, $KE = W_p + W_f$ $\frac{1}{2}mv^2 = F_p\cdot (21.0~m) - F_f\cdot (10.0~m)$ $v^2 = \frac{(2)(225~N)(21.0~m)-(2)(36.0~kg)(9.80~m/s^2)(0.20)(10.0~m)}{36.0~kg}$ $v^2 = 223.3~m^2/s^2$ $v = \sqrt{223.3~m^2/s^2} = 14.9~m/s$ The final speed of the crate is 14.9 m/s.
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