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

`F_f=840N`

Step-by-step 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`