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JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Self Evaluation Test - Work, Energy and Power

  • 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\[\operatorname{F}\left( x \right)=-\operatorname{kx}+a{{x}^{3}}\]. 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] \[{{\operatorname{dU}}_{\left( x \right)}}=-Fdx\] \[\therefore \,\,\,{{U}_{x}}=-\int_{0}^{x}{Fdx=\frac{k{{x}^{2}}}{2}}\frac{a{{x}^{4}}}{4}\] \[U=0\,at\text{ }x=0\text{ }and\text{ }at\text{ }x=\sqrt{\frac{2k}{a}};\Rightarrow \]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 \[F=0.\,But\,F=-\frac{dU}{dx}\] \[\Rightarrow \operatorname{At}\text{ }x=0;\text{ }\frac{dU}{dx}=0\]i.e. the slope of the graph should be zero. These characteristics are represented by [d].


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