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Calculate the Work

  1. Dec 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Calculate the work performed by the force field F = <y,x,z^3> from P = (1,1,1) to Q = (2,3,4).




    2. Relevant equations



    3. The attempt at a solution
    So first I found that F must be a conservative field because ∇(xy + (z^4)/r) = <y,x,z^3>

    Since it is a conservative field it is path independent... which means something that my teacher would be mad at me for not knowing (lol)

    Can I parameterize the line integral in terms of t?

    like
    c(t) = <1+t,1+2t,1+3t>
    c'(t) = <1,2,3>

    ∫F dot c'(t) dt = W? then what would be the bounds on my integral?
     
  2. jcsd
  3. Dec 9, 2013 #2

    haruspex

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    Since you've found the potential function, why not just take the potential difference?
     
  4. Dec 9, 2013 #3
    ^what he or she said. Which is, coincidentally, a very useful result of path independence!
     
    Last edited: Dec 9, 2013
  5. Dec 9, 2013 #4

    LCKurtz

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    What values of ##t## correspond to the two points?
     
  6. Dec 10, 2013 #5
    0 and 1?
    so ∫F dot c'(t) dt = W between 0 and 1 will give me the correct answer?
     
  7. Dec 10, 2013 #6
    how do I put F into terms of t though? like x = cos(t) y = sin(t) and z=z? ;-( i'm confused
     
  8. Dec 10, 2013 #7

    LCKurtz

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    Why do you answer every hint with another question? Aren't you capable of plugging ##t=0## and ##t=1## into your equation of the line and checking whether you get the two points yourself?

    Yes, if you insist on doing the problem the hard way. Had you heeded the advice in post #2 you would have been done with this thread long ago.
     
  9. Dec 10, 2013 #8
    I got the work = 68.75.

    f(r(Q) - f(r(P)
    f = xy + (z^4)/4
    from <1,1,1> to <2,3,4>
    so
    (2)(3) + ((4)^4)/4 - ((1)(1) + ((1)^4)/4
    70 - 1.25
    68.75

    Is this the correct method for doing this problem?
     
  10. Dec 10, 2013 #9

    haruspex

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    Yes. (Btw, you wrote /r instead of /4 in the OP. Adjacent keys, I see.)
     
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