JEE Main & Advanced AIEEE Solved Paper-2004

  • 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     AIEEE  Solved  Paper-2004  

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

    B)        \[{{e}^{x}}\]                      

    C)        \[x\]                     

    D)        \[{{\log }_{e}}x\]

    Correct Answer: A

    Solution :

    In this problem acceleration (a) is given in terms of displacement\[(x)\]to determine the velocity with respect to position or displacement we have to apply integration method. From given information\[a=-\text{ }kx,\]where a is acceleration,\[x\]is displacement and\[k\]is a proportionality constant. \[\frac{v\,dv}{dx}=-kx\]                \[\left[ \because \frac{dv}{dt}=\frac{dv}{dx}\left( \frac{dx}{dt} \right)=\frac{v\,dv}{dx} \right]\] \[\Rightarrow \] \[v\,dv=-k\,xdx\] 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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