Calculate the Kinetic Frictional Force

[SRooney]

The alarm at a fire station rings and a 71.3-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 3.63 m). Just before landing, his speed is 1.42 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?

Givens:
m = 71.3 kg
d = 3.63 m
Vf = 1.42 m/sd = Vf(t)/2
a = Vf/t
F = ma

3.63 m = 1.42 m/s (t) /2
7.26 m = 1.42 m/s (t)
t = 5.113 s

a =1.42 m/s /5.113 s
a = .278 m/s^2

F = 71.3 kg (.278 m/s^2)
F = 19.8 N

But that's not the answer

 

[cnh1995]

The alarm at a fire station rings and a 71.3-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 3.63 m). Just before landing, his speed is 1.42 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?

Givens:
m = 71.3 kg
d = 3.63 m
Vf = 1.42 m/sd = Vf(t)/2
a = Vf/t
F = ma

3.63 m = 1.42 m/s (t) /2
7.26 m = 1.42 m/s (t)
t = 5.113 s

a =1.42 m/s /5.113 s
a = .278 m/s^2

F = 71.3 kg (.278 m/s^2)
F = 19.8 N

But that's not the answer

I didn't check your calculations but the force you have calculated is the "net force" on the fireman's body. You are asked to find the frictional force.