Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 10 - Interactions and Potential Energy - Exercises and Problems - Page 259: 56


The speed of the block is 3.3 m/s

Work Step by Step

We can find the work done by the force as; $W = \int_{0}^{4.0}~F_x~dx$ $W = \int_{0}^{4.0}~(20-5x)~dx$ $W = (20x-2.5x^2)~\Big\vert_{0}^{4.0}$ $W = 40~J$ The kinetic energy will be equal to the sum of the work done by the force and the work done by friction; $KE = W+W_f$ $\frac{1}{2}mv^2 = W-mg~\mu_K~d$ $v^2 = \frac{2W-2mg~\mu_K~d}{m}$ $v = \sqrt{\frac{2W-2mg~\mu_K~d}{m}}$ $v = \sqrt{\frac{(2)(40~J)-(2)(2.6~kg)(9.80~m/s^2)(0.25)(4.0~m)}{2.6~kg}}$ $v = 3.3~m/s$ The speed of the block is 3.3 m/s.
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