1. The problem statement, all variables and given/known data A skier of mass 70.0kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him a distance of 60.0m up a 30.0 degree slope (assumed frictionless) at a constant speed of 2.00m/s? [tex] Given:[/tex] [tex] \Delta r = 60m [/tex] [tex] \theta = 30^\circ [/tex] [tex] Mass = 70kg [/tex] 2. Relevant equations [tex] W=F\Delta r\cos(\theta) [/tex] 3. The attempt at a solution I've solved the problem, but not the way I'd like to. You may notice that, with all given data, that the triangle drawn from this would be a 30/60/90, which lets you solve it without using the work formula. You could also do it with potential / kinetic energy. This is how my horrible (would require its own thread; it's cal based physics, he doesn't teach any of the calculus aspect) physics professor did it, not bothering to showing us how to solve it with the work formula, which is what the point of the problem was, as that's what the chapter is on. When he gave us the work formula, he wrote it on the board and gave us one problem dealing with a level plane, said the [tex]\cos(\theta)[/tex] would be 1 because the angle was [tex]90^\circ[/tex] (he didn't explain this), and said to ignore it; that's it. Nobody in my class knows how to deal with incline planes using the work formula. This angers the hell out of me because I want to major in physics, and I'm going to have to end up setting myself back a semester by either passing this class and auditing/retaking a better professor's class on a branch campus, or taking this grade and teaching myself. After reading online, I found a spot where someone got a force by doing [tex]\sin(30^\circ)(70kg)[/tex], cutting the weight to [tex]35kg[/tex] (they said because the slope supports some of the weight). They proceeded to use the work formula, but without the [tex]\cos(\theta)[/tex] part. The final answer ends up being [tex]20580J[/tex], but I don't know WHY. Long story short, can someone PLEASE explain this to me? I would be forever grateful.