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MGIMS WARDHA MGIMS WARDHA Solved Paper-2015

  • question_answer
    The potential energy of a certain spring when stretched through a distance\[x\]is 20 J. The amount of work that must be done on this spring to stretch it through an additional distance \[\frac{x}{2}\]will be

    A)  20J                        

    B)  25J  

    C)  5J                                          

    D)  15J

    Correct Answer: B

    Solution :

                    Initial potential energy\[=\frac{1}{2}k{{x}^{2}}=20J\] Final potential energy\[=\frac{1}{2}k{{\left( x+\frac{x}{2} \right)}^{2}}\]                 \[=\frac{9}{4}.\frac{1}{2}k{{x}^{2}}=\frac{9}{4}\times 20=45J\] So, work done = change in potential energy of spring \[=45-20=25J\]


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