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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 4

  • question_answer
    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 \[F(x)=-kx+a{{x}^{3}}\]. Here k and a are positive constants. For \[x\ge 0,\] the functional form of the potential energy U(x) of the particle is

    A)

    B)

    C)

    D)

    Correct Answer: D

    Solution :

    \[d{{U}_{(x)}}=-Fdx\]
    \[\therefore \,\,{{U}_{x}}=-\int_{0}^{x}{Fdx}=\frac{k{{x}^{2}}}{2}-\frac{a{{x}^{4}}}{4}\]
    \[U=0\]at \[x=0\] and at \[x=\sqrt{\frac{2k}{a}};\,\Rightarrow \] we have potential energy zero twice (out of which one is at origin).
     Also, when we put \[x=0\]in the given function, we get \[F=0\]. But \[F=-\frac{dU}{dx}\]
    \[\Rightarrow \] At \[x=0;\] \[\frac{dU}{dx}=0\] i.e. the slope of the graph should be zero. These characteristics are represented by [d].


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