An elevator's cable snaps and its emergency brakes start gripping at the secondary line in order to slow its descent. The elevator and passengers weigh 389 kgs together and fall under the influence of gravity (9.81 m/s2). The coefficient of friction between the brakes and secondary line is 0.8, while the emergency brakes are designed such that the normal force on the line is equal to the weight of the elevator and passengers. How fast is the elevator going after falling 8 stories? (One story is precisely 4.50 meters)
Blah the gravity and friction is confusing me can someone help?
Mass of elevator & passengers, m = 389 kg
g=9.81 m/s²
Normal force on the brakes, N
= weight of elevator and passengers
= mg
Fricitonal force
= μN
Net downward acceleration, a
= mg - μN
Height of 8 storeys, H
= 8 *4.5 m
= 36 m
Initial velocity, v0 = 0 m/s
Final velocity, v
then
v² - u² = 2aH
Solve for v.
This is what I done before I end up getting v=26.5887 and for some reason it's wrong =(. I not sure why though.
How did you get 26.6 m/s?
Acceleration du to gravity is g=9.81 m/s²
The frictional resistance force
= μmg
= 0.8 mg
Therefore the net downward force
= mg - 0.8mg
= 0.2mg
Net downward acceleration
= 0.2g
I think 26.6 m/s comes from
v=sqrt(2aH)
where a=9.81 m/s² instead of 0.2g.