Work by Gravity / Conservative Forces

AI Thread Summary
The discussion centers on proving that gravity is a conservative force using a specific diagram and equations. The work done (W) in moving a ball through various points is calculated, showing that the total work done around a closed path is zero. The sign convention indicates that work is negative during upward movement and positive during downward movement, reflecting how gravitational potential energy converts to kinetic energy. Clarification is provided on the height variable, emphasizing that different elevations must be considered in calculations. The conversation concludes with an understanding of the relationship between height and work done by gravity.
Jamieee
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A ball is moved from point A to point B, to point C, to point D, to point E, and back to point A. So, to make things easier... here's a diagram of the problem.

http://img181.imageshack.us/img181/125/conservativeforcesvy3.jpg

We were asked to prove (mathematically) the concept that gravity (in this diagram) is a conservative force, and why the angle in the diagram is disregarded.

The following equations were given (and I don't really understand them :confused:)...

W = W (A to B) + W (B to C) + W (C to D) + W (D to E) + W (E to A)
W = 0 + (-mgh) + 0 + (-mgh) + (mgh)
W = 0 <--- This has to be proven as well.

I'm also wondering why W (D to E) became (-mgh) instead of Fxcos(theta).

Help please... Thanks in advance!

--- Jamie
 
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W = W (A to B) + W (B to C) + W (C to D) + W (D to E) + W (E to A)
W = 0 + (-mgh) + 0 + (-mgh) + (mgh)
It appears the sign convention is - for upward movement and - for downward movement, i.e. the if ball ascends then gravity does - work, and if the ball descends then gravity is doing + work. If the ball were in freefall (downward) the gravitational potential energy would be transformed into the balls kinetic energy which would increase.

Please be careful with the distance h. One has used h repeatedly, but there are different elevations or heights involved.

If h is the distance AE, then BC < h. The height between D and E is DE sin\theta, and then h= AE = BC + DE sin\theta.
 
Thank you. So much. I get it now. :D
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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