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

  • question_answer A spring of force constant 10 N/m has an initial stretch 0.20 m. In changing the stretch to 0.25 m, the increase in potential energy is about                                  [CPMT 1977]

    A)             0.1 joule

    B)               0.2 joule

    C)             0.3 joule

    D)               0.5 joule

    Correct Answer: A

    Solution :

                    DP.E.\[=\frac{1}{2}k(x_{2}^{2}-x_{1}^{2})=\frac{1}{2}\times 10[{{(0.25)}^{2}}-{{(0.20)}^{2}}]\]             \[=5\times 0.45\times 0.05=0.1\,J\]

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