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Manipal Medical Manipal Medical Solved Paper-2004

  • question_answer
    300 J of work is done in sliding a 2 kg block up an inclined plane of height 10m. The work done against friction is (Take\[g=10m/{{s}^{2}}\])

    A)  zero        

    B)  100 J

    C)  200 J       

    D)  300 J

    Correct Answer: B

    Solution :

     Given: Work done in sliding a block up an inclined plane W = 320 Joule Mass of block M = 2 kg Height of inclined plane = 10 m The potential energy stored in the block is given by \[U=mgh=2\times 10\times 10=200\,J\] Hence, work done against friction is given by \[=W-U=300-200=100\,J\]


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