- #1
lalalah
- 18
- 0
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?
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 = (v2y - v2oy)/(2ay), where ay can be substituted with ay= (-fk)/mass
Sooo, my messy and essentially incorrect attempt:
y = -3.55 m
v0 = 0 m/s
voy = 1.79 m/s
mass = 96.0 kg
using the equation for the y-displacement, I attempted to find fk:
-3.55 m = (1.792 - 0 m/s) / (2(-fk)/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!
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 = (v2y - v2oy)/(2ay), where ay can be substituted with ay= (-fk)/mass
The Attempt at a Solution
Sooo, my messy and essentially incorrect attempt:
y = -3.55 m
v0 = 0 m/s
voy = 1.79 m/s
mass = 96.0 kg
using the equation for the y-displacement, I attempted to find fk:
-3.55 m = (1.792 - 0 m/s) / (2(-fk)/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!