A fireman of mass 60 , kg slides down a pole. | vedclass

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(D) The fireman is sliding down the pole. The forces acting on him are his weight $(W = mg)$ acting downwards and the kinetic frictional force $(f_k =

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$5$

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(D) The fireman is sliding down the pole. The forces acting on him are his weight $(W = mg)$ acting downwards and the kinetic frictional force $(f_k = \mu R)$ acting upwards,where $R$ is the normal force applied by the fireman on the pole.Given:Mass $m = 60 \, kg$Normal force $R = 600 \, N$Coefficient of friction $\mu = 0.5$Acceleration due to gravity $g = 10 \, m/s^2$The net force acting on the fireman in the downward direction is $F_{net} = mg - f_k$.Using Newton's second law,$F_{net} = ma$,we have:$ma = mg - \mu R$$a = g - \frac{\mu R}{m}$Substituting the values:$a = 10 - \frac{0.5 	imes 600}{60}$$a = 10 - \frac{300}{60}$$a = 10 - 5 = 5 \, m/s^2$Thus,the fireman slides down with an acceleration of $5 \, m/s^2$.

$A$ 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 (in $, m/s^2$)? (Take $g = 10 \, m/s^2$)

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