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 9 - Work and Kinetic Energy - Exercises and Problems: 52

Answer

$W = (0.632)~L~F_0$

Work Step by Step

We can find the work that the force does on the particle as it moves from x = 0 to x = L $W = \int_{0}^{L}~F_x~dx$ $W = \int_{0}^{L}~F_0~e^{-x/L}~dx$ $W = -L~F_0~e^{-x/L}~\Big\vert_{0}^{L}$ $W = (-L~F_0~e^{-L/L})-(-L~F_0~e^{-0/L})$ $W = (-L~F_0~\frac{1}{e})-(-L~F_0)$ $W = L~F_0~(1-\frac{1}{e})$ $W = (0.632)~L~F_0$
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