Coefficient of kinetic friction on a slide with inclined and level sections

[songokou77]

When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 1.89 m/s. The mass stops a distance S2 = 1.85 m along the level part of the slide. The distance S1 = 1.25 m and the angle theta = 37.7 degrees. Calculate the coefficient of kinetic friction for the mass on the surface.

Well i kinda found a formula for this by changing the conservatiion of energy(K_f+U_f=K_i+U_i+W_nc).

The formula is mu_k=(-1/2*(V_i^2)-1/2*(g*h))/(g*S_total),

But I still don't get the right answer am I doing something wrong?

 

[arildno]

Think again:
You haven't used the information that you are on an INCLINED plane!

 

[songokou77]

I used the incline plane to figure out h=Ssin37.7=0.26

 

[arildno]

It's still wrong, you've used a frictional force: $$\mu{mg}$$
This is incorrect.
In addition, your signs are messed up