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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    A particle is placed at the origin and a force \[F=k\text{ }x\] is acting on it (where k is positive constant). If u (0) = 0, the graph of \[u(x)\] versus \[x\] will be (where u is potential energy function) :

    A)                                                

    B)

    C)                                                             

    D)

    Correct Answer: D

    Solution :

                     Given, \[F=kx\]where k is positive constant. Force,                   \[F=\frac{-dU(x)}{dx}\] or                            \[\frac{dU(x)}{dx}=-k\,x\] or                            \[dU(x)=-kx\,\,dx\] Integrating both the sides, \[U(x)=-\int_{{}}^{{}}{kx\,dx=-\frac{1}{2}}\,k{{x}^{2}}\] or                            \[{{x}^{2}}=-\frac{2}{k}U(x)\]                     ?(i) Comparing Eq. (i) with the standard equationof parabola, which is given by \[{{x}^{2}}=4ay\] we conclude that the graph of \[U(x)\]versus \[x\]satisfies the graph of parabola. Since the equation (i) has minus sign,therefore the graph will be open downwardsalong \[U(x).\]


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