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JIPMER Jipmer Medical Solved Paper-2012

  • question_answer
    A stone weighing 1 kg and sliding on ice with a velocity of 2 m/s is stopped by friction in 10 s. The force of friction (assuming it to be constant) will be

    A)  - 20 N            

    B)         - 0.2 N

    C)  0.2 N             

    D)         20 N

    Correct Answer: B

    Solution :

    Here, initial velocity u = 2 m/s final velocity v = 0 Time duration t = 10 s \[a=\frac{v-u}{t}=\frac{0-2}{10}=\frac{-2}{10}=\frac{-1}{5}\] \[\therefore \] Friction force F = ma = \[F=ma=1\times (-\,0.2)=-0.2\,N.\]


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