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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2013

  • 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)  100J

    C)  zero                     

    D)  1000J

    Correct Answer: A

    Solution :

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


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