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## Homework Statement

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?

## Homework Equations

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!