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JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Graphical Questions

  • 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 from of the potential energy \[U(x)\] of the particle is              [IIT-JEE (Screening) 2002]            

    A)

    B)

    C)

    D)      

    Correct Answer: D

    Solution :

                                \[F=\frac{-dU}{dx}\Rightarrow dU=-F\,\,dx\]             \[\Rightarrow U=-\int_{0}^{x}{(-Kx\,+\,a{{x}^{3}})dx}\]\[=\frac{k{{x}^{2}}}{2}-\frac{a{{x}^{4}}}{4}\]             \[\therefore \] We get U = 0 at x = 0 and x = \[\sqrt{2k/a}\]             and also U = negative for \[x>\sqrt{2k/a}\].             So F = 0 at x = 0             i.e. slope of U ? x graph is zero at x = 0.


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