J & K CET Engineering J and K - CET Engineering Solved Paper-2014

  • question_answer
    An object is gently placed on a long converges belt moving with \[11\text{ }m{{s}^{-1}}\]. If the coefficient of friction is \[0.4,\] then the block will slide in the belt up to a distance of

    A)  \[10.21\,\,m\]          

    B)  \[15.43\,\,m\]

    C)  \[20.3\text{ }m\]            

    D)  \[25.6\text{ }m\]

    Correct Answer: B

    Solution :

    \[u=11m/s\] \[a=mg\] \[=0.4\times 10=4m/{{s}^{2}}\] \[v=0\] So from \[{{v}^{2}}={{u}^{2}}+2as\] \[0={{(11)}^{2}}+2(-4)\times s\] \[s=\frac{11\times 11}{8}=15.125m\]


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