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

  • question_answer Natural length of a spring is 60 cm, and its spring constant is 4000 N/m. A mass of 20 kg is hung from it. The extension produced in the spring is, (Take \[g=9.8\ m/{{s}^{2}}\]) [DCE 2004]

    A)             4.9 cm

    B)             0.49 cm

    C)               9.4 cm

    D)             0.94 cm

    Correct Answer: A

    Solution :

                    If x is the extension produced in spring.             \[F=kx\] Þ \[x=\frac{F}{k}\]= \[\frac{mg}{k}=\frac{20\times 9.8}{4000}=4.9\,cm\]


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