Energy Conservation of an object on an incline

[inner08]

An 1kg object is sliding down a 4m high incline which is inclined at 53 degrees. The initial speed is 2m/s. It then slides on a horizontal section 3m long which is at ground level and then slides up an incline plane that is inclined at 37 degrees. All surfaces have a kinetic coefficient of uc = 0.4. What distance will the object travel up the 37 degree incline before stopping?

Work Done:

Starting off Values:
Vi = 2m/s
Vf = ?
d = 3.19 (using trig)
m = 1kg
uc = 0.4

I drew a diagram and then I figured I would use the laws of conservation formula. I know their is kinetic and potential energy when the object is at a certain height. When the object is on the horizontal surface, their will only be kinetic and friction forces acting on it. So I thought i'd find the object's final speed on the 53degree incline (1/2mVf^2 = 1/2mVi^2 + mgh -fx). After I found Vf, I used that as the new initial speed for the horizontal surface. I then thought i'd find the final speed on the horizontal speed using the laws of conservation of energy (1/2mVf^2 = 1/2mVi^2 -fx). Finally, I took this value and used it as the new initial speed as the object is about to move up the 37 degree incline. I substituted it in the formula: mgh = 1/2mVi^2 - fx.

I keep getting an answer like 2.41m but the answer in the book is 1.95m. I'm not quite sure where I went wrong?

 

[neutrino]

Your idea is right, you may want to re-check you calculations. Also, h is not the required answer - it's h/sin(37).