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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Potential Energy
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kane (2179)

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the P>E> for a conservative system is u=ax2-bx.what is the P.E. at the pos. of equilibrium
 
ans:-b2/4a

there are numerous options besides I.I.T

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rhd92781 (686)

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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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Greatdreams (3083)

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The best way to approach this sort of a problem is to

draw the graph......


If you draw the graph it comes out to be a parabola

With axes as Energy and Displacement

I dont know how to draw the graphs here but it can be

easily found that the latus rectum comes out to be

-b^2/4a which is the potential energy at equilibrium.



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kane (2179)

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thnxs a lot both of you

there are numerous options besides I.I.T

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<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>









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