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Question

A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x)=kx+ax3. Here k and a are positive constants. For x0, the functional form of the potential energy U(x) of the particle is


A


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B


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C


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D


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Solution

The correct option is D



The potential energy of the particle is given by
U=Fdx=(kx+ax3)dx
U=kx22ax44=x24(2kax2).......(1)
From Eq. (1) it follows that U = 0 at two values of x which are x = 0 and x=2ka. Hence, graphs (b) and (c) are not possible.
Also, U is maximum or minimum at a value of x given by dUdx=0
ddx(kx22ax44)=0kxax3=0
x=ka
At this value of x, U is maximum if d2Udx2<0.
Now, d2Udx2=ddx(kxax3)=k3ax2
At x=ka, d2Udx2=k3aka
=k3k=2k, which is negative.
Hence U is maximum at x=ka.
Hence, graph (a) is also not possible.
Moreover, U is negative for x>2ka.
Therefore, the correct graph is (d).


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