Roger pushes a box on a 30° incline. if he applies a force of 60 newtons parallel to the incline and displaces the box 10 meters along the incline, how much work will he do on the box?

Pretty sneaky, I must say !  

                   Work = (force) x (distance)

That formula says nothing about the directions of the force OR
the distance.  As long as they're both in the same direction, they
can both be horizontal (sliding a crate on the floor), vertical (lifting
something), or on any slant in between.

Roger's force and distance are both in the same direction ... the
direction of the incline ... so

                   Work = (60 newtons) x (10 meters)  =  600 joules.

Kiloleter, mega leterhope this !