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Punjab Medical Punjab - MET Solved Paper-2003

  • question_answer
    If a body of mass \[3\,\,kg\] is dropped from the top of a tower of height \[25\,\,m\], then kinetic energy after \[3\sec \] will be:

    A) \[557\,\,\text{J}\]                           

    B) \[246\,\,\text{J}\]

    C) \[1048\,\,\text{J}\]                        

    D) \[1297\,\,\text{J}\]

    Correct Answer: D

    Solution :

    Here: mass\[m=3\text{ }kg\], height\[h=25m\] and  time\[t=3\sec \] Final velocity of body after\[3\sec \],                 \[v=u+gt=0+9.8\times 3=29.4\,\,m/s\] Hence, kinetic energy of the body is given by                 \[=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\times 3\times {{(29.4)}^{2}}\]


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