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Manipal Engineering Manipal Engineering Solved Paper-2010

  • question_answer
    A body is coming with a velocity of 72 km/h on a rough horizontal surface of coefficient of friction 0.5. If the acceleration due to gravity is \[10\,m/{{s}^{2}},\] find the minimum distance it can be stopped.

    A)  400m                                   

    B)  40m

    C)  0.40m                  

    D)         4m

    Correct Answer: B

    Solution :

    Given,\[u=72km/h=20m/s\] \[a=\mu g=0.5\times 10\,\,m/{{s}^{2}}\] From     \[{{v}^{2}}={{u}^{2}}-2as\] \[\therefore \]  \[{{(0)}^{2}}={{(20)}^{2}}-2\times 0.5\times 10\times s\] \[\therefore \]  \[s=\frac{20\times 20}{2\times 0.5\times 10}\] or            \[s=40\,\,m\]


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