JCECE Medical JCECE Medical Solved Paper-2004

  • question_answer
    A child is swinging a swing. Minimum and maximum heights of swing from earths surface are 0.75 m and 2m respectively. The maximum velocity of this swing is:

    A)  5 m/s        

    B)  10 m/s

    C)  15 m/s       

    D)  20 m/s

    Correct Answer: A

    Solution :

     Key Idea: Maximum kinetic energy of swing should be equal to difference in potential energies to conserve energy. From energy conservation                  \[\frac{1}{2}mv_{\max }^{2}=mg({{H}_{2}}-{{H}_{1}})\] Here,\[{{\text{H}}_{\text{1}}}\text{=}\]maximum height of swing from earths surface \[=0.75\,m\] \[{{H}_{2}}=\]maximum height of swing from earths     surface \[=2\,m\] \[\therefore \] \[\frac{1}{2}mv_{\max }^{2}=mg(2-0.75)\] or \[{{v}_{\max }}=\sqrt{2\times 10\times 1.25}=\sqrt{25}=5\,m/s\]


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