An elevator in a tall building is allowed to reach a maximum speed of 3.3 m/s going down. What must the tension be in the cable to stop this elevator over a distance of 3.4 m if the elevator has a mass of 1330 kg including occupants?

Answer 1

First calculate the acceleration "a" needed to decelerate in that distance Y.
V = sqrt(2aX)
a = V^2/(2X) = 1.6 m/s^2

The tension T in the cable during deceleration is :
T = M (g + a)

Answer 2

5.6
n