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: 26b


$U_{\max }=U_{0}+12\times 2-3\times 2^{2}=39J$

Work Step by Step

Let's 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 $12x-3x^{2}$ must have its maximum value becouse $U_{0} $ is constant so firstly lets calculate maximum value of $12x-3x^{2}$ İn order to calculate $12x-3x^{2}=3\left( 4x-x^{2}\right) =-3\left( x^{2}-4x\right) =-3\left( x^{2}-4x+4\right) +12=12-3\left( x-2\right) ^{2} $ we see here it will be maximum when $x=2$ so maximum ptential energy will be $U_{\max }=U_{0}+12\times 2-3\times 2^{2}=39J$
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