The way I see if we could lift the crate straight up in method 1, no ramp.
We know what h is. It is 3 meters * sin A
Then the work done = m g h = 3 m g sin A
In method 2 we do twice as much work as in method 1. However the work we do against gravity in method 2 remains mgh so the work done against friction is also mgh
the work done against friction is the same as mgh so
3 m g sin A = 485 J
3(45)(9.8) sin A = 485
sin A = .367
A = 21.5 degrees
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 45.0-kg crate straight up(method 1) or push it up a 3.00-m-long 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.
I've posted this question, and someone gives me this answer. But I don't quite get it, can anyone please explain it to me?Thanks!
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 = ?
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