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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Second Law of Motion

  • question_answer
    The average resisting force that must act on a 5 kg mass to reduce its speed from 65 cm/s to 15 cm/s in 0.2s is [RPET 2000]

    A)             12.5 N

    B)               25 N

    C)             50 N   

    D)             100 N

    Correct Answer: A

    Solution :

                    \[F=m\,\left( \frac{v-u}{t} \right)=\frac{5(65-15)\times {{10}^{-2}}}{0.2}=12.5\,N\]


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