Work done on a block on an inclined plane

[mariners02]

1. Homework Statement [/b]
A block of weight mg sits on an inclined plane as shown. A force of magnitude F is applied to pull the block up the incline at constant speed. The coefficient of kinetic friction between the plane and the block is Mu.
What is the total work done on the block by the applied force as the block moves a distance up the incline?
Express your answer in terms of any or all of the variables Mu, m, g, theta, L and F

Homework Equations

W=FDcos(theta)

The Attempt at a Solution

It seems like since W=FDcos(theta) the answer would simply be FLcos(theta) but its telling me that is wrong, can anyone give me an idea of what I'm missing. Thanks!

 

[superdave]

1. Homework Statement [/b]
A block of weight mg sits on an inclined plane as shown. A force of magnitude F is applied to pull the block up the incline at constant speed. The coefficient of kinetic friction between the plane and the block is Mu.

The Attempt at a Solution

It seems like since W=FDcos(theta) the answer would simply be FLcos(theta) but its telling me that is wrong, can anyone give me an idea of what I'm missing. Thanks!

You are missing the part I put in bold

 

[AdrianMay]

The answer is FL. Who cares what's opposing the force.

 

[mariners02]

FL was right, thanks adrianmay, why wouldn't it be FLcos(theta)?

 

[RTW69]

Theta is not the angle of the incline it is the angle between the direction of the force and the direction of travel. Since the force and direction of travel are parallel the angle between them is 0. Cos[0] = 1 therefore W=FLcos[0] equals FL

 

[AdrianMay]

Exactly. Just look at the force and the direction it's moving. The rest is irrelevant.