An elevator in a tall building is allowed to reach a maximum speed of 3.89 m/s going down. What must the tension be in the cable to stop this elevator over a distance of 2.70 m if the elevator has a mass of 1320 kg including occupants?
Vfy^2 = Viy^2 + 2ay(Yf-Yi)
F = ma (Newton's Second Law in the y-direction)
The Attempt at a Solution
Vi = 3.89
T = ?
Yf = 2.70 m
Yi = 0
m = 1320 kg
a = ?
Using the first equation: 0 = (3.89)^2 + 2(ay) (2.70)
ay = -2.80 m/s^2
Applying Newton's Second Law, I got: T - mg = ma
T = mg + ma = (1320)(9.81) + (1320)(-2.80) = 9250.24 N
However, the system says it's incorrect. Any help would be appreciated!