Question

The alarm at a fire station rings and a 84.5-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 4.47 m). Just before landing, his speed is 1.49 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?

Answer 1

`F_k = 807.82 N`

Step-by-step explanation:

As we know that fireman starts from rest

so here we have

`v_i = 0`

`v_f = 1.49 m/s`

`y = 4.47 m`

now we can use kinematics to find the acceleration

`v_f^2 - v_i^2 = 2 a y`

`1.49^2 - 0 = 2(a)(4.47)`

`a = 0.25 m/s^2`

as we know by force equation

`mg - F_k = ma`

`(84.5)(9.81) - F_k = 84.5(0.25)`

`F_k = (84.5)(9.81 - 0.25)`

`F_k = 807.82 N`