Work done on an inclined slope?

[quincyboy7]

Homework Statement

Imagine a block with mass m being pulled up a frictionless incline with upward angle theta by a rope with tension force T. The mass is pulled until another point of rest at a point distance d away from the original point (with d measured on the incline) and h measured on the y-axis from the ground. What is the work done by the gravity force and the tension force?

Homework Equations

W=-U
W=delta K=Fdcostheta

The Attempt at a Solution

Alright so the work done by gravity is: W=mgdcos(theta+90)=-mgdsin*(theta)=-mgh.
This can also be derived by W=-U with gravity. Since the mass is at rest in both positions, delta K is 0 and Wnet is also 0. Thus, W done by tension is just equal to mgh.

My question is about direction, the gravity force's work is done opposite to the intended motion, and the tension's force in the same direction. I get that. However, the mgh interpretation of the work done by gravity leads to the interpretation all energy is added only in the vertical direction since gravity is a vertical force? I don't see how this can coincide with the fact that the work magnitude is equal for gravity is equal to that of tension, since gravity is directed straight down when tension is directed on the incline itself. The block is certainly displaced in the direction of the slope, but does the fact that it comes to rest with some element of gravity counteracting tension play into the net force/work discussion? I am completely confused conceptually. Any clear explanation would be greatly appreciated.

 

[Delphi51]

Work is not a vector quantity; it has no sense of direction.
Another way to think of the question is that, since there is no friction, the route doesn't matter. You could pull the mass along the floor beside the ramp (no work), then lift the mass straight up to height h and the work would be mgh. Just my 2 cents worth.