Posted by ajo on Sunday, December 9, 2007 at 4:23pm.
At the end of the semester, you are loading your gear into a van to go home. In loading the van, you can either lift a 45.0kg crate straight up(method 1) or push it up a 3.00mlong ramp that is inclined at an angle above the horizontal (method 2).In both methods, the crate is moved at a constant speed and the vertical distance from the ground to the floor of the van has the same value h. In method 1, no friction is present. In method 2, however, a frictional force does 485J of work on the crate. In applying your nonconservative pushing force to the crate, you do twice as much work in method 2 as in method 1. Find the angle.
Can anyone please give me some hints to do this?THANKS A LOT!

physics  drwls, Sunday, December 9, 2007 at 5:39pm
First, compute the work done in method 1. That would be M g H = 45*9.8*3.0 sin A = 1323 sin A Joules. A is the angle of the place to the horizontal.
With the ramp, you do twice as much work, or 2646 J sin A. If 485 J of this was overcoming friction, the potential energy change was
2646 sin A  485
This equals the potential energy change with method 1, or 1323 sin A. Therefore
2646 sin A 485 = 1323 sin A
485 = 1323 sin A
sin A = 0.367
A = ? 
physics  ajo, Sunday, December 9, 2007 at 6:02pm
i don't understand...this part:
the potential energy change was
2646 sin A  485
This equals the potential energy change with method 1, or 1323 sin A. Therefore
2646 sin A 485 = 1323 sin A
485 = 1323 sin A
can you please explain it in details£¿