Energy conservation of a sliding box

[utsharpie]

3.4kg rock starts at 25 cm on a plane with a 33degree angle. upon reaching the bottom the box slides along the horizontal. friction coeff is
.19. how far does the box slide on the horizontal before coming to a rest?

ok i need to use .5MVi2 + MGy1 = .5MVf2 + MGy2 + Ffrd

first i need to find the velocity of the box as it hits the end of the incline. i also need the length of the incline.

i have Y1=.025 meters
Vf=0
Vi=0
M=3.4kg
g=9.8
theta=33
fr=.19

i keep getting 4.3 meters but its wrong

 

[utsharpie]

ok i found the distance from top of incline to bottom of incline to be .459meters using laws of triangles. so if friction is acting the velocity is gsin33-.19gcos33 and that equals 3.7758

v2=2ad
v2=3.466
v=1.8618

ok so we got Vi at the bottom to be 1.861 and the y1 and y2 at the bottom are both zero and the final V is 0. so is it mv2/2=.19d? and that comes out to 31.01425 meters?no way!