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CMC Medical CMC-Medical VELLORE Solved Paper-2010

  • question_answer
    300 J of work is done in sliding up a 2 kg block on an inclined plane to a height of 10 m. Taking value of acceleration due to gravity g to be 10 \[m{{s}^{-2}},\]work done against friction is

    A)  100 J                                    

    B)  200 J

    C)  300 J                                    

    D)  zero

    E)  None of these

    Correct Answer: B

    Solution :

                    Work done against gravity = mgh \[=2\times 10\times 10=200\,J\] Work done against friction = (Total work done - work done           against gravity) \[=300-200=100\text{ }J\]


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