University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 7 - Potential Energy and Energy Conservation - Problems - Exercises - Page 230: 7.28


(a) W = 0 (b) W = -0.104 J (c) W = 0.104 J (d) The force is conservative. $U = (4~N/m^2) ~x^3$

Work Step by Step

(a) Since the force is directed toward the (-x)-axis, the force does zero work on the proton because the force is at a $90^{\circ}$ angle to the path of motion. (b) $W = \int_{0.10}^{0.30}F~dx$ $W = \int_{0.10}^{0.30}-\alpha ~x^2~dx$ $W = \frac{-\alpha}{3} ~x^3\vert_{0.10}^{0.30}$ $W = (-4~N/m^2)~((0.30~m)^3-(0.10~m)^3)$ $W = -0.104~J$ (c) $W = \int_{0.30}^{0.10}F~dx$ $W = \int_{0.30}^{0.10}-\alpha ~x^2~dx$ $W = \frac{-\alpha}{3} ~x^3\vert_{0.30}^{0.10}$ $W = (-4~N/m^2)~((0.10~m)^3-(0.30~m)^3)$ $W = 0.104~J$ (d) As the proton moves from (0.10 m, 0) to (0.30 m, 0) and back to the same point, the total work done by the force on the proton is -0.104 J + 0.104 J, which is zero. Therefore the force is conservative. $U = U_0 - \int F~dx$ $U = 0 - \int -\alpha ~x^2~dx$ $U = -\frac{-\alpha}{3} ~x^3$ $U = (4~N/m^2) ~x^3$
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