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J & K CET Medical J & K - CET Medical 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. 21m.          

    B)  15.43m

    C)  20.3m            

    D)  25.6m

    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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