J & K CET Medical J & K - CET Medical Solved Paper-2003

  • question_answer
    A particle moves along the x-axis from \[x={{x}_{1}}\]to \[x={{x}_{2}}\] under the action of a force given by F = 2x. Then the work done in the process is :

    A)  zero                                     

    B)  \[x_{2}^{2}-x_{1}^{2}\]

    C) \[2{{x}_{2}}({{x}_{2}}-{{x}_{1}})\]                            

    D) \[2{{x}_{1}}({{x}_{1}}-{{x}_{2}})\]

    Correct Answer: B

    Solution :

                    Mechanical work (W) is the force applied through a distance, defined mathematically as the line integral of a scalar product of force (F) and displacement\[(dx)\]vectors. \[W=\int{Fdx}\] Given, \[F=2x,\] Using    \[\int{{{x}^{n}}}dx=\frac{{{x}^{n+1}}}{n+1}\] \[\therefore \]  \[W=\int_{{{x}_{1}}}^{{{x}_{2}}}{2x}dx=2\left[ \frac{{{x}^{2}}}{2} \right]_{{{x}_{1}}}^{{{x}_{2}}}\]                 \[=(x_{2}^{2}-x_{1}^{2})\]


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