The alarm at a fire station rings and a 96.0-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 3.55 m). Just before landing, his speed is 1.79 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?(adsbygoogle = window.adsbygoogle || []).push({});

Relevant equations:

I'm not sure, but I believe this has something to do with Newton's second law of motion? So this may be helpful {[tex]\Sigma[/tex]Fnet }/mass = acceleration , although I couldn't see a way to use it myself

I mainly used: y = (v^{2}_{y}- v^{2}_{oy})/(2a_{y}), where a_{y}can be substituted with a_{y}= (-f_{k})/mass

3. The attempt at a solution

Sooo, my messy and essentially incorrect attempt:

y = -3.55 m

v_{0}= 0 m/s

v_{oy}= 1.79 m/s

mass = 96.0 kg

using the equation for the y-displacement, I attempted to find f_{k}:

-3.55 m = (1.79^{2}- 0 m/s) / (2(-f_{k})/96kg)

multiplying both sides by -(2/96) fk ----> .07396 fk = 1.79^2

fk = 43.3 N

if someone could point me in the right direction i would be terribly grateful!

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# Homework Help: Newton's Second Law of Motion, Kinetic Friction Force of Fireman Sliding Down a Pole

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