RAJASTHAN PMT Rajasthan - PMT Solved Paper-2010

  • question_answer
    When a spring is stretched by 2 cm, it stores 100 J of energy. If it is stretched further by 2 cm, the stored energy will be increased by

    A)  100 J                                    

    B)  200 J

    C)  300 J                    

    D)         400 J

    Correct Answer: C

    Solution :

    The energy of spring                 \[100=\frac{1}{2}k{{x}^{2}}\]                   \[W=\frac{1}{2}k(x_{2}^{2}-x_{1}^{2})=\frac{1}{2}k[{{(2x)}^{2}}-{{x}^{2}}]\]                        \[=3\times \left( \frac{1}{2}k{{x}^{2}} \right)=3\times 100\] If the spring is stretched further by\[2\,\,cm\], the stored energy will be increased by\[300\,\,J\].


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