A fireman of mass 60, kg slides down a pole. He is pressing pole with a force of 600 ,N . coefficien

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4-2.Friction

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A fireman of mass $60\, kg$ slides down a pole. He is pressing the pole with a force of $600 \,N$. The coefficient of friction between the hands and the pole is $0.5$, with what acceleration will the fireman slide down ........ $m/s^2$

A$1$

B$2.5$

C$10$

D$5$

Solution

(d) Net downward acceleration $ = \frac{{{\rm{Weight – Friction \,force}}}}{{{\rm{Mass}}}}$
$ = \frac{{(mg – \mu \;R)}}{m}$
$ = \frac{{60 \times 10 – 0.5 \times 600}}{{60}}$
$ = \frac{{300}}{{60}} = 5\;m/{s^2}$

Standard 11

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(JEE MAIN-2022)

A fireman of mass $60\, kg$ slides down a pole. He is pressing the pole with a force of $600 \,N$. The coefficient of friction between the hands and the pole is $0.5$, with what acceleration will the fireman slide down ........ $m/s^2$

Solution

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