Question

The alarm at a fire station rings and a 87.5-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 4.10 m). Just before landing, his speed is 1.75 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?

Answer 1

Answer:

$F_f=840N$

Explanation:

From the question we are told that

Weight of fireman $W_f= 87.5kg$

Pole distance $D=4.10m$

Final speed is $V_f 1.75m/s$

Generally the equation for velocity is mathematically represented as

$v^2 = v_0^2 + 2 a d$

Therefore Acceleration a

$a'= v^2 / 2 d$

$a'= 0.21m/s^2$

Generally the equation for Frictional force $F_f$ is mathematically given as

$F_f=m*a$

$F_f=m*(g-0.21)$

$F_f=87.5*(9.81-0.21)$

Therefore

$F_f=840N$