JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति JEE PYQ - Work Energy Power and Collision

  • question_answer
    A block of mass m, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant k. The other end of the spring is fixed, as shown in the figure. The block is initally at rest in its equilibrium position. If now the block is pulled with a constant force F, the maximum speed of the block is-[JEE Main 09-Jan-2019 Morning]

    A)             \[\frac{2F}{\sqrt{mk}}\]

    B) \[\frac{\pi \,F}{\sqrt{mk}}\]

    C)             \[\frac{F}{\pi \,\sqrt{mk}}\]

    D) \[\frac{F}{\sqrt{mk}}\]

    Correct Answer: D

    Solution :

    [d]
               
                For maximum speed \[F\text{ }=\text{ }kx\]
                \[x\,\,=\,\,\frac{f}{k}\]
                \[Fx\,\,=\,\,\frac{1}{2}k{{x}^{2}}+\frac{1}{2}m{{V}^{2}}\]
                \[\frac{{{F}^{2}}}{k}\,\,=\,\,\frac{1}{2}\,\,\frac{k{{F}^{2}}}{{{k}^{2}}}\,\,+\,\,\frac{1}{2}\,m{{V}^{2}}\]
                \[\frac{{{F}^{2}}}{2k}\,\,=\,\,\frac{m{{V}^{2}}}{2}\]
                \[V\,\,=\,\,\frac{F}{\sqrt{mk}}\]


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