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Posted by 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 45.0-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.

Can anyone please give me some hints to do this?THANKS A LOT!

  • physics - , 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 - , 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£¿

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