Work on an incline with friction.

[d.tran103]

Okay, I am stuck on this problem that comes in two parts. Please help! Thanks!

Homework Statement

Part 1) Daphne pushes a 22.0 kg crate up a frictionless 30.0 degree incline a distance along the incline of 10.0 m at constant velocity. How much work did Daphne do?

Part 2) Daphne now pushes a 22.0 kg crate up the 30.0 degree incline with a coefficient of kinetic friction of 0.200. As before, she pushes the crate a distance up the incline of 10.0 m at constant velocity. How much work did Daphne do?

Homework Equations

W=Fdcos(theta)

The Attempt at a Solution

Part 1) I'm 100% confident in the answer I got for part 1, w=(9.8 m/ss)(22 kg)(10.0 m)(cos60), w=1.08E3 J.
Part 2) This is where I'm stuck. How does friction apply to this? It changes the force but it isn't working for me. I know the answer is 1451.4J but can't figure out how to get it. Thanks!

 

[gomunkul51]

Force of kinetic friction of the moving body = coefficient of kinetic friction * normal force from the surface on the body.
Direction of the force of friction is against the direction of movement.

 

[d.tran103]

So is 431.2 N my diagonal component and 215.6 my vertical component, making 215.6 the normal force? Thanks!