Haryana PMT Haryana PMT Solved Paper-2008

  • question_answer
    300 J of work is done in sliding a 2 kg block up     an inclined plane of height 10 m. Taking \[g=10m/{{s}^{2}}\] 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                 \[\Rightarrow \]               \[W={{W}_{g}}+{{W}_{f}}\]             ??(i) but         \[W=300J\]                 \[{{W}_{g}}=mgh\]                 \[=2\times 10\times 10=200J\]  Putting in Eq. (i), we get                 \[300=200+{{W}_{f}}\] \[\Rightarrow \]               \[{{W}_{f}}=300-200=100J\]


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