Work done by gravity on an inclined plane

[Mr Davis 97]

I know that since gravity is a conservative force, the work done by it to displace an object some distance does not depend on the path taken. This leads to the conclusion that the work done by gravity in sliding an object down an inclined plane of height h would be $mgh$. However, what is the significance of $(mg\sin\theta)\Delta d$, where Δd is the distance traveled along the hypotenuse? This is not the work done by gravity, but what does the quantity represent?

 

[brainpushups]

Well, isn't that also just the work done when friction is absent? Write it as
$$mg (h/d) d$$
and you end up with
$$mgh$$

Think of it as the work done by the component of gravity along the direction of the plane.