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: 61


(a) $F = -\frac{B~\pi}{L}$ (b) $F = -AL$ (c) $F = -2AL+\frac{B~\pi}{L}$

Work Step by Step

We can find an expression for $F$: $U = Ax^2+B~sin~(\frac{\pi x}{L})$ $F = -\frac{dU}{dx} = -2Ax-\frac{B~\pi}{L}~cos~(\frac{\pi x}{L})$ (a) We can find $F$ when $x = 0$: $F = -2Ax-\frac{B~\pi}{L}~cos~(\frac{\pi x}{L})$ $F = -2A(0)-\frac{B~\pi}{L}~cos~(0)$ $F = -\frac{B~\pi}{L}$ (b) We can find $F$ when $x = \frac{L}{2}$: $F = -2Ax-\frac{B~\pi}{L}~cos~(\frac{\pi x}{L})$ $F = -2A(\frac{L}{2})-\frac{B~\pi}{L}~cos~(\frac{\pi}{2})$ $F = -AL$ (c) We can find $F$ when $x = L$: $F = -2Ax-\frac{B~\pi}{L}~cos~(\frac{\pi x}{L})$ $F = -2A(L)-\frac{B~\pi}{L}~cos~(\pi)$ $F = -2AL+\frac{B~\pi}{L}$
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