VMMC VMMC Medical Solved Paper-2010

  • question_answer
    A block of mass 0.5 kg has an initial velocity of \[10\,m{{s}^{-1}}\] down an inclined plane of angle friction between the block and the inclined surface is 0.2 The velocity of the block after it travels a distance of 10 m is

    A)  \[17\,m{{s}^{-1}}\]

    B)                         \[13\,m{{s}^{-1}}\]

    C)  \[24\,m{{s}^{-1}}\]        

    D)         \[8\,m{{s}^{-1}}\]

    Correct Answer: B

    Solution :

                            Here, \[m=0.5\,kg,u=10\,m{{s}^{-1}},\theta ={{30}^{o}}\] \[\mu =0.2,s=10m.\] Acceleration down the plane, \[a=g(sin\theta -\mu \cos \theta )\]  \[=10(sin{{30}^{o}}-0.2cos{{30}^{o}})=3.268\,m{{s}^{-2}}\] From,   \[{{v}^{2}}={{u}^{2}}+2as\]                 \[={{10}^{2}}+2(3.268)\times 10\] \[=165.36\] \[v=\sqrt{165.36}\approx 13\,m{{s}^{-1}}\]


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