A particle, which is constrained to move along the x− axis, is subjected to a force from the origin as F(x)=−kx+ax3. Here k and a are positive constants. For x=0, the functional form of the potential energy U(x) of 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 Slope of potential energy vs x graph is force.
Given, F(x)=−kx+ax3;F(x)=0;
for x=0x=√ka;
So, slope is zero at x=0x=√ka U=−∫Fdx U=∫(kx−ax3)dx U=(2kx2−ax4)4+C
As x increases beyond √ka potential energy contiuously reduces
only option possible is d