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Question

A boy of mass m is sliding down a vertical pole by pressing it with a horizontal force f. If μ is the coefficient of friction between his palms and the pole, the acceleration with which he slides down will be equal to


A

g

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B

μfm

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C

g+μfm

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D

gμfm

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Solution

The correct option is D

gμfm


Here, normal reaction, R=f
Therefore, force of friction, fr=μR=μf
Net downward force, F=mgfr=mgμf
From Newton's second law of motion, acceleration, a=Fm=mgμfm=gμfm
Hence, the correct choice is (d).


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