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**1. The Problem Stament, all variables and given data**

a 15 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20 degrees with the horizontal. if there is a constant force of kinetic friction of 25 N between the crate and ramp, what kinetic energy would the crate have at the bottom of the ramp?

**2. Needed equations**

Work = PE + KE + Energy Lost

PE= mgh

Height = Sinθ * Length

KE= (1/2) m*v^2

**3. Attempt at solution**

okay so I know how to start out with finding PE

PE= mass* gravity* height = 15 * 9.8 * height

Find Height

Height= Sin (20°) * 2.0= .684

PE= 15 * 9.8 * .684 = 100.5 Joules

What's throwing me off is the kinetic friction of 25 N, I'm not sure what to do with it and where to plug it into to