The alarm at a fire station rings and a 87-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 3.9 m). Just before landing, his speed is 1.3 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?

Answer 1

The frictional heat loss F*X equals the loss of mechanical energy

M g H - (1/2)M*Vfinal^2 = F*X

M cancels out Solve for F

H = 3.9 m
g = 9.8 m/s^2
Vfinal = 1.3 m/s