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Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: Potential Energy
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[Post New]15 Jan 2008 23:12:19 IST
    Subject: Re:Potential Energy
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rhd92781 (686)

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Olaaa!! Perrrfect answer. 118  [166 rates]
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its simple
F=-dU/dx
=-d(ax^2-bx)/dx
=b-2ax
At equilibrium, F=0
b-2ax=0
x=b/2a
so, PE = a(b/2a)^2 - b(b/2a)
= -b2/4a
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