VMMC VMMC Medical Solved Paper-2013

  • question_answer
    When a spring is stretched by a distance \[x,\]it  exerts a force given by \[F=(-5x-16{{x}^{3}})N\] The work done, when the spring is streached from 0.1m to 0.2m is

    A) \[8.7\times {{10}^{-2}}J\]

    B)  \[12.2\,\times {{10}^{-2}}J\]

    C)  \[8.1\times {{10}^{-1}}J\]

    D)  \[12.2\times {{10}^{-1}}J\]

    Correct Answer: A

    Solution :

     \[F=-(-5x-16{{x}^{3}})N\] \[F=-(5+16{{x}^{2}})x\] Restoring force \[F=-kx\] \[k=5+16{{x}^{2}}\] \[\omega =\frac{1}{2}{{k}_{2}}x_{2}^{2}-\frac{1}{2}{{k}_{1}}x_{1}^{2}\] \[{{x}_{1}}=0.1\,m\]and \[{{x}_{2}}=0.2\,m\] \[\omega =\frac{1}{2}[5+16{{(0.2)}^{2}}]{{(0.2)}^{2}}\] \[-\frac{1}{2}[5+16{{(0.1)}^{2}}]{{(0.2)}^{2}}\] \[=2.82\times 4\times {{10}^{-2}}-2.58\times {{10}^{-2}}\] \[=8.7\times {{10}^{-2}}J\]


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