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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2011

  • question_answer
    Block A of mass of 2 kg is placed over block B of mass 8 kg. The combination is placed over a rough horizontal surface. Coefficient of friction between B and the floor is 0.5. Coefficient of friction between blocks A and B is 0.4. A horizontal force of 10 N is applied on block B. The force of friction between blocks A and B is \[(g=10\,m{{s}^{-2}})\]

    A)  100 N                                   

    B)  40 N

    C)  50 N                                     

    D)  Zero

    Correct Answer: D

    Solution :

    Total mass of blocks A and \[B=2+8=10\text{ }kg\]Friction between surface and combination of A and B                 \[F=\mu R\]                 \[=0.5\times 10\times 10=50\,N\] Here applied force on box B is ION that is less than 50 N. So the system will be in rest because of this there is no friction between blocks A and B.


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