Work on an incline with friction.

AI Thread Summary
Daphne pushes a 22.0 kg crate up a 30.0 degree incline in two scenarios: first on a frictionless incline and then with kinetic friction. In the first part, she calculates the work done as 1.08E3 J using the formula W=Fdcos(theta). For the second part, she struggles to incorporate the effect of kinetic friction, which changes the force required to maintain constant velocity. The expected answer for the work done against friction is 1451.4 J, but she is confused about calculating the normal force and frictional force. Understanding the relationship between these forces is crucial for solving the problem accurately.
d.tran103
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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!
 
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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.
 
So is 431.2 N my diagonal component and 215.6 my vertical component, making 215.6 the normal force? Thanks!
 

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