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

  • question_answer When a 1.0kg mass hangs attached to a spring of length 50 cm, the spring stretches by 2 cm. The mass is pulled down until the length of the spring becomes 60 cm. What is the amount of elastic energy stored in the spring in this condition, if g = 10 m/s2                           [MP PET 2001]

    A)             1.5 Joule

    B)               2.0 Joule

    C)             2.5 Joule

    D)               3.0 Joule

    Correct Answer: C

    Solution :

                    Force constant of a spring             \[k=\frac{F}{x}=\frac{mg}{x}=\frac{1\times 10}{2\times {{10}^{-2}}}\] Þ \[k=500\,N/m\]             Increment in the length = 60 ? 50 = 10 cm             \[U=\frac{1}{2}k{{x}^{2}}=\frac{1}{2}500\,{{(10\times {{10}^{-2}})}^{2}}=2.5\,J\]

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