Calculating Work on a Ramp: 20kg Crate with Constant Force and Friction

[victor.raum]

Homework Statement

A 20kg create sits at rest at the bottom of a 15m long ramp that is inclined at 34deg above the horizontal. A constant horizontal force is of 290N is applied to the crate to push it up the ramp. While the crate is moving a constant frictional force of 65N is exerted by the ramp in resistance to its motion. What is the total work done on the crate during its motion from the bottom to the top of the ramp?

Homework Equations

W=F \cdot d

The Attempt at a Solution

I found the component of force parallel to motion to be

F_{||} = F_x \cdot \cos\left(\frac{34}{360} \cdot 2\pi\right)

One can then find the work by subtracting the 65N of friction resistance and then multiplying by the distance of the motion, like so:

W = \left(F_{||} - 65N\;\;\right)\cdot d

When I plug in the numbers I get 2631 Joules, but my answer key says it's 987 Joules. Can someone clue me in?

 

[Doc Al]

Don't forget gravity.

 

[victor.raum]

Doh, right; yep, that comes out to 987 Joules now.

Somehow I saw the force of gravity that pressed the block up against the ramp and the resulting normal force as canceling and thus rendering both irrelevant. Totally missed that there is (now obviously) also a portion of the weight vector which does not press up against he ramp, since the ramp is slanted and not laying flat.

Thanks a million.