JEE Main & Advanced Physics Work, Energy, Power & Collision Question Bank Work Done by Variable Force

  • 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                        [Orissa JEE 2002]

    A)             100 J  

    B)             200 J

    C)             300 J

    D)             400 J

    Correct Answer: C

    Solution :

                    \[100=\frac{1}{2}k{{x}^{2}}\]                  (given)             \[W=\frac{1}{2}k(x_{2}^{2}-x_{1}^{2})=\frac{1}{2}k[{{(2x)}^{2}}-{{x}^{2}}]\]             \[=3\times \left( \frac{1}{2}k{{x}^{2}} \right)=3\times 100=300\,J\]      

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