JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

  • question_answer
    A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10s. Then, the coefficient of friction is [AIEEE 2003]

    A)       \[0.02\]

    B) \[0.03\]

    C) \[0.06\]

    D) \[0.01\]

    Correct Answer: C

    Solution :

    [c] Let coefficient of friction be \[\mu \], then retardation will be \[\mu \]g.
    From equation of motion, v = u + at, we get
                \[0=6-\mu g\times 10\]                (\[\because a=\mu g\])
                \[\Rightarrow \]   \[\mu =\frac{6}{100}=0.06\]

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