VMMC VMMC Medical Solved Paper-2007

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

    A) 200 J                     

    B)        100 J                     

    C) zero                      

    D)        1000 J

    Correct Answer: B

    Solution :

    Net work done in sliding a body up to a height h on inclined plane = Work done against gravitational force + Work done against frictional force \[\Rightarrow \]               \[W={{W}_{g}}+{{W}_{g}}\]                                        ?(i) but         \[W=300\,J\] \[{{W}_{g}}=mgh=2\times 10\times 10=200\,J\] Putting in Eq. (i) we get \[300=200+{{W}_{f}}\] \[{{W}_{f}}=300-200=100\,\text{J}\]


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