JCECE Engineering JCECE Engineering Solved Paper-2002

  • question_answer
    A long spring is stretched by \[2\,\,cm\]. Its potential energy is \[U\]. If the spring is stretched by \[10\,\,cm,\] the potential energy stored in it will be:

    A) \[\frac{U}{25}\]                               

    B) \[\frac{U}{5}\]

    C) \[3U\]                                  

    D)  \[25U\]

    Correct Answer: D

    Solution :

    When a spring is stretched by \[x\,\,cm\], its stored potential energy is                 \[U=\frac{1}{2}k{{x}^{2}}\] where \[k\] is spring constant. Given,   \[{{U}_{1}}=U,\,\,{{x}_{1}}=2\,\,cm,\,\,{{x}_{2}}=10\,\,cm\] \[\therefore \]  \[\frac{{{U}_{1}}}{{{U}_{2}}}=\frac{x_{1}^{2}}{x_{2}^{2}}\] \[\Rightarrow \]               \[\frac{U}{{{U}_{2}}}=\frac{{{(2)}^{2}}}{{{(10)}^{2}}}\] \[\Rightarrow \]               \[{{U}_{2}}=\frac{100}{4}U=25\,\,U\] Note: On stretching the spring further potential energy increases.


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