VMMC VMMC Medical Solved Paper-2010

  • question_answer
    A spring of force constant 800 Nm1 has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is

    A)  16 J       

    B)                         8 J         

    C)  32 J                       

    D)         24 J

    Correct Answer: B

    Solution :

                    Work done, \[W=\frac{1}{2}k(x_{2}^{2}-x_{1}^{2})\] \[=\frac{1}{2}\times 800[{{(15\times {{10}^{-2}})}^{2}}-{{(5\times {{10}^{-2}})}^{2}}]\] \[W=400\times 200\times {{10}^{-4}}=8\,\,\text{J}\]


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