- #1

SRooney

Member warned to use the homework template for posts in the homework sections of PF.

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/s

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

Givens:

m = 71.3 kg

d = 3.63 m

Vf = 1.42 m/s

**d = Vf(t)/2**

a = Vf/t

F = maa = 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