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

  • question_answer The force constant of a wire is k and that of another wire is \[2k.\] When both the wires are stretched through same distance, then the work done                           [MH CET 2000]

    A)             \[{{W}_{2}}=2W_{1}^{2}\]

    B)               \[{{W}_{2}}=2{{W}_{1}}\]

    C)             \[{{W}_{2}}={{W}_{1}}\]

    D)               \[{{W}_{2}}=0.5{{W}_{1}}\]

    Correct Answer: B

    Solution :

                    \[W=\frac{1}{2}k{{x}^{2}}\]             If both wires are stretched through same distance then \[W\propto k\]. As \[{{k}_{2}}=2{{k}_{1}}\] so \[{{W}_{2}}=2{{W}_{1}}\]


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