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Manipal Medical Manipal Medical Solved Paper-2000

  • question_answer
    A cylinder of 10 kg is rolling in a plane with an initial velocity of 10 m/s. If the coefficient of friction between the surface and cylinder is 0.5 kg then before slipping, it will cover:\[(g=10m/{{s}^{2}})\]

    A)  2.5m            

    B)  5m

    C)  7.5m            

    D)  10m

    Correct Answer: D

    Solution :

     Here: Mass of the cylinder \[m=10\text{ }kg\] Initial velocity of cylinder \[v=10\text{ }m/s\] Coefficient of friction between surface and the cylinder \[\mu =0.5\] The kinetic energy of the cylinder is \[=\frac{1}{2}m{{\upsilon }^{2}}\] \[=\frac{1}{2}\times 10\times {{10}^{2}}\] \[=500kgm/sec\]         ...(1) The work done by the cylinder in rolling over the plane is = Force of friction\[\times \]distance \[=\mu mg\times s\] \[=0.5\times 10\times 10\times s\] \[=50s\]                      ...(2) As the kinetic energy of the cylinder will be consumed in doing the work. Hence, from (1) and (2) \[500=50s\] \[s=\frac{500}{50}=10\,m\]


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