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

  • question_answer A spring with spring constant k when stretched through 1 cm, the potential energy is U. If it is stretched by 4 cm. The potential energy will be                        [Orissa PMT 2004]

    A)             4U      

    B)             8U      

    C)             16 U   

    D)             2U

    Correct Answer: C

    Solution :

                    Potential energy \[U=\frac{1}{2}k{{x}^{2}}\]             \\[U\propto {{x}^{2}}\][if k = constant] If elongation made 4 times then potential energy will become 16 times.

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