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 257: 33

Answer

At $x=2~m$: $F_x = 2.5~N$ At $x=5~m$: $F_x = 0.40~N$ At $x=8~m$: $F_x = 0.16~N$

Work Step by Step

$U = \frac{10}{x}~J$ We can find an expression for $F_x$: $F_x = -\frac{dU}{dx}$ $F_x = -(-\frac{10}{x^2})~N$ $F_x = \frac{10}{x^2}~N$ We can find $F_x$ at $x = 2~m$: $F_x = \frac{10}{x^2}~N$ $F_x = \frac{10}{2^2}~N$ $F_x = 2.5~N$ We can find $F_x$ at $x = 5~m$: $F_x = \frac{10}{x^2}~N$ $F_x = \frac{10}{5^2}~N$ $F_x = 0.40~N$ We can find $F_x$ at $x = 8~m$: $F_x = \frac{10}{x^2}~N$ $F_x = \frac{10}{8^2}~N$ $F_x = 0.16~N$
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