Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 8 - Potential Energy and Conservation of Energy - Problems - Page 204: 26a


$U\left( x\right) =U_{0}+12x-3x^{2}=27+12x-3x^{2}$

Work Step by Step

Lets assume that our starting point is $x=0$ so the work done by conservative force at any given $x$ will be $W=\int ^{x}_{0}F\left( x\right) dx=\int ^{x}_{0}\left( 6x-12\right) =3x^{2}-12x(1)$ So the change in potential energy for conservative force will be: $\Delta U=-W\left( 2\right) $ So from (1) and (2) we get $\Delta U=-W=12x-3x^{2}(3)$ And also $\Delta U=U\left( x\right) -U_{0}\left( 4\right) $ From (3) and (4) we get: $U\left( x\right) =U_{0}+12x-3x^{2}$ in order $U(x)$ to be maximum positive $12x-3x^{2}$
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