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## Homework Statement

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?

## Homework Equations

If I'm thinking correctly, I'd use the work equation...

W=ΔK+ΔUg+ΔUs

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∅)

Then...

(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?