A .444kg block, slides down a .888m ramp, at an angle of 33.3°. The block starts from rest, and reaches a speed of 2.22 m/s. How much energy is lost?
If I'm thinking correctly, I'd use the work equation...
ΔK=(1/2)m*vf^(2) - (1/2)m*vi^(2) (The second part of this drops out, because the initial velocity is 0)
Oh, and there's no spring involved, so ΔUs drops out.
The Attempt at a Solution
So, the stated equation can be re-written as...
(1/2)m*vf^2 +m*g*Δy (ΔY=Δxsin∅)
(1/2)(.444kg)(2.22^(2)m/s^(2)) + (.444kg)(9.80m/s^2)(.888m sin(33.3°))
The answer I get, assuming I'm right is 3.22 J? Am I getting anywhere with this, or am I completely off?