JCECE Medical JCECE Medical Solved Paper-2008

  • question_answer A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement  \[x\] is proportional to

    A) \[{{x}^{2}}\]

    B)  \[{{e}^{x}}\]

    C)  \[x\]

    D)  \[{{\log }_{e}}x\]

    Correct Answer: A

    Solution :

     From given information \[a=-\,kx,\]where a is acceleration, \[x\] is displacement and \[k\] is a proportionality constant. \[\frac{v\,dv}{dx}=-\,k\,x\] \[\Rightarrow \] \[v\,dv=-\,k\,x\,dx\] Let for any displacement from 0 to \[x,\] the velocity changes from \[{{v}_{0}}\]to v. \[\Rightarrow \] \[\int_{{{v}_{0}}}^{v}{v\,dv=-\int_{0}^{x}{k\,x\,dx}}\] \[\Rightarrow \] \[\frac{{{v}^{2}}-v_{0}^{2}}{2}=-\frac{k\,{{x}^{2}}}{2}\] \[\Rightarrow \] \[m\left( \frac{{{v}^{2}}-v_{0}^{2}}{2} \right)=-\frac{mk\,{{x}^{2}}}{2}\] \[\Rightarrow \] \[\Delta K\propto {{x}^{2}}\] [\[\Delta K\]is loss in KE]


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