Work by Gravity / Conservative Forces

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SUMMARY

The discussion centers on proving that gravity is a conservative force through the analysis of work done as a ball moves along a specified path. The key equations presented include W = W (A to B) + W (B to C) + W (C to D) + W (D to E) + W (E to A) and the simplification leading to W = 0. The sign convention indicates that gravity does negative work when the ball ascends and positive work when it descends. Additionally, the importance of accurately defining the height 'h' in relation to different elevations is emphasized.

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  • Understanding of conservative forces in physics
  • Familiarity with work-energy principles
  • Knowledge of gravitational potential energy equations
  • Basic trigonometry for analyzing angles and heights
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  • Study the mathematical proof of conservative forces in physics
  • Learn about gravitational potential energy and its applications
  • Explore the concept of work done by forces in different scenarios
  • Investigate the implications of sign conventions in physics problems
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Students of physics, educators teaching mechanics, and anyone interested in understanding the principles of conservative forces and gravitational work.

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[itex]\theta[/itex], and then h= AE = BC + DE sin[itex]\theta[/itex].
 
Thank you. So much. I get it now. :D
 

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