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Homework Help: Force fields, vectors, and work (mostly just confused by notation)

  1. Apr 21, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider a force field F = c(iy - jx). From the force field calculate the work required to move a particle from the origin to the point 2i + 4j without acceleration along the two different paths:

    • From the origin to 2i then to 2i + 4j
    • From the origin to 4j then to 2i + 4j

    Comment whether the force field is conservative or not

    2. The attempt at a solution

    I'm just slightly confused by the notation used in the question. I know that i and j are just the unit vectors in the x and y directions, but what is c in the force field expression? And if i and j are already noted in this expression, why are x and y used as well?

    Assuming c is just some arbitrary constant, then is the work done simply 6c J for both paths? My logic for this is that for the first path the work done is 2c J along the x-axis and then 4c J along the y-axis (and the other way around for the second path).

    Thus the force field is conservative (work done is independent of the path taken).

    Is my logic correct, or am I missing something? Thank you :smile:
  2. jcsd
  3. Apr 21, 2010 #2


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    Homework Helper

    There are conditions for the conservativeness of a force field, you should check one of these (formally) before concluding if it is conservative. If c is an arbitrary constant, x and y are probably variables, so F=F(x, y).
  4. Apr 21, 2010 #3
    I don't understand - how can I check these conditions formally? Have I not already done so by showing that the work done is the same for both paths in the first part of the question? Note that I am asked to comment on whether or not the force field is conservative, which implies that there are no additional calculations required.
  5. Apr 21, 2010 #4


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    Well, then that's it - you have shown that the work is independent of the path taken.

    You could also calculate ∇ x F, which equals 0 for a conservative force field, but since you're not asked to..
  6. Apr 21, 2010 #5
    OK, thanks! :)
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