Roller coaster problem involving velocity

[y90x]

Homework Statement

A roller coaster reaches the top of the steepest hill with a speed of 7.0 km/h. It then descends the hill, which is at an average angle of 43° below horizontal and is 46.0 m long. What will its speed be when it reaches the bottom? Assume the coefficient of kinetic Friction is 0.12.

Homework Equations

Ff= U•Fn
F=ma
Vf ^2 - Vo^2 = 2ad

The Attempt at a Solution

Ff=U•Fn
Ff= U•mgcos(43)

F=ma
Mgsin(43) - Umgcos(43) =ma
The mass cancels out here so ,
gsin(43) - Ugcos(43) = a
5.82 = a

Vf= (square roof of>>) 2ad + vo ^2
Vf = (square roof of >> ) 2 • 5.82 • 0.046 + 7.0^2
Vf = 7.04 km/h

The answer should be 83.6 km/h
Where did I go wrong

 

[TSny]

5.82 = a

What are the units for this acceleration?

 

[y90x]

What are the units for this acceleration?

I just realized it was m/s^2
Thank you