Tension in elevator cable stopping from 3.89 m/s over 2.70 m

[PhysicsCCR]

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!

 

[ehild]

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

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

Think over the signs. Considering the upward direction positive, the displacement of the elevator is negative, -2.7 m. The acceleration points upward, it is positive.

ehild

 

[PhysicsCCR]

Oh, okay. Thank you, Ehild!