What is the Total Energy of a Rolling Sphere on a Sloped Floor?

[Extremist223]

Homework Statement

A solid metal ball of radius 0.500m, and a mass of 1.50 kg, is found to be rolling down a sloped floor whose angle is 30.0° to the horizontal (assume no slipping). The ball has a final angular velocity of 2.00 rad/s. What is the total energy of the ball when it it is 2.00m from the bottom of the slope? (Assume there is no friction)

Homework Equations

Energy final = 1/2mv^2 + 1/2Iw^2 + mghf
v= rw
v= .5 (2) = 1m/s
I = moment of inertia = 2/5(mr^2)

The Attempt at a Solution

Ef= 1/2(1.5kg)(1m/s^2)+ 1/2(2/5(1.5kg(.5m^2) + 1.5kg(9.8m/s^2)(2m)
Ef= 0.75 + 0.075 + 29.4
Ef= 30.225 Joules

Does this seem right?

 

[mgb_phys]

I suspect they mean a distance of 2m along the slope to the bottom
what is the 'h' in the potential energy formula

 

[Extremist223]

thank you, that makes the answer right it ends up being sin30 (2m)= 1 which makes my answer right thanks a lot.