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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Motion on Inclined Surface

  • question_answer
    300 Joule of work is done in sliding up a 2 kg block on an inclined plane to a height of 10 metres. Taking value of acceleration due to gravity ?g? to be 10 m/s2, work done against friction is                              [MP PMT 2002]

    A)             100 J  

    B)             200 J

    C)             300 J  

    D)             Zero

    Correct Answer: A

    Solution :

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


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