Calculate speed of a hoop up an incline

[yoslick11]

Homework Statement

Ok, I have attempted this problem, and believe I have the right answer. My work is not matching up though and I think I am missing something.

Question:
A hoop is rolling without slipping along a horizontal surface with a speed of 4.50 m/s when it starts up a ramp that makes and angle of 25.0° with the horizontal. What is the speed of the hoop after it has rolled 3.00m up the ramp?

Homework Equations

KE = 1/2*mv^2 + 1/2*I*w^2
I = mR^2

The Attempt at a Solution

This is what I got.

Total KE = 1/2*mv^2+1/2*Iw^2
substituting the I = mR^2 and w = v/R

Total KE = 1/2*m*v^2 + 1/2*(mR^2)*(v^2/R^2)
= 1/2*m*v^2 + 1/2*m*v^2
= mv^2

at 3m up on the incline

ke(bot) = pe(top)+ke(top)

m*v(initial)^2 = mgh + 1/2*m*v(final)^2
vi^2 = gh + 1/2*vf^2

vf = sqrt(2(vi^2-gh))

I believe at this point I am wanting vf = sqrt(vi^2-gh).. because the answer I believe is 2.80 m/s. sqrt(4.5^2 - (9.8*3sin25)) = 2.80 m/s

Am I looking at this correctly?

Any help would be appreciated.

 

[rcgldr]

You calculated that total KE for the rolling hoop is m v², and used m v² for ke(bot) but then you used 1/2 m v² for ke(top)?