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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    A constant force F is-pushing a 5 kg mass on a horizontal surface at a constant velocity of 2 m/s. The coefficient of friction between the surface and the mass is 0.3 (Take\[g=10\text{ }m/{{s}^{2}}\]). If F acts along the direction of motion, the rate at which F is doing work (in watt):

    A)  3                            

    B)         6

    C)  10                         

    D)         15

    E)  30

    Correct Answer: E

    Solution :

    \[Power=Fv=\mu mgv\] Given, \[\mu =0.3,g=10m/{{s}^{2}},m=5\,kg,v=2m/s\] \[\therefore \]  \[P=0.3\times 5\times 10\times 2=30\,W\]


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