JEE Main & Advanced Sample Paper JEE Main Sample Paper-4

  • question_answer
    A boy of mass m is sliding down a vertical pole by pressing it with a horizontal force F. If u. is the coefficient of friction between his palm & the pole, the acceleration with which he slides down is

    A)  1                                            

    B) \[\frac{\mu F}{m}\]

    C) \[g+\frac{\mu F}{m}\]                  

    D) \[g-\frac{\mu F}{m}\]

    Correct Answer: D

    Solution :

    Here normal contact force N = F friction force, \[f=\mu N=\mu F\] So the FBD of boy is ma = mg-f\[\Rightarrow \]\[a=g-\frac{\mu F}{m}\]

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