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Work done on a particle by a nonconservative force

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle moves from a point of origin on an x,y plane to the point (5,5) with the units of the plane being meters. The force the particle experiences is given by the vector < 2y, x^2 >. Calculate the work done on the particle as it moves from (0,0) to (5,5).



    2. Relevant equations
    So I know work done is equal to the integral of the dot product of the force and displacement vectors.


    3. The attempt at a solution
    So I attempted to take the integral from (0,0) to (5,5) of < 2y , x^2 > * < dx, dy> and got an answer of 175 J. But the solution is apparently 66.7 J. I have no idea how to arrive to such a solution.

    Much help would be appreciated, thanks.

    Edit: Okay, I can see how to arrive at such a solution if it were the dot product of < 2y, x^2> * < dy, dx >. Is this simply an error on the textbook's part? Having < dy, dx > does not make sense.
     
  2. jcsd
  3. Oct 18, 2009 #2

    rl.bhat

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    Work done by Fx = Intg(x^2)dx between 0 to 5.
    Work done by Fy = Intg(2y)dy between 0 to 5
    Add them to get net work done.
     
  4. Oct 18, 2009 #3
    But why is it (x^2)dx and (2y)dy when the x component is 2y and the y component is x^2?
     
  5. Oct 18, 2009 #4

    rl.bhat

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    Because Fx.dx = (2y).dx = 0 So to reach the point (5 , 5), we have to take Fx.dy and Fy.dx
     
  6. Oct 18, 2009 #5
    Does that only apply then the path of trajection is along <dx,dy>? Because a previous question related to this problem was to find the work done when going along the x-axis 5 m to the right and then going up to the point (5,5). In that case I did integral from (0,0) to (5,0) of 2ydx plus integral from (5,0) to (5,5) of (x^2)dy. I did so because the displacement vector for the first integral is < dx , 0 > and the displacement vector for the second integral is < 0 , dy >, and my answer was correct.
     
  7. Oct 18, 2009 #6

    rl.bhat

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    Work done does not depend on the path taken by particle. It only depends on the initial position and the final position. So we have a choice to take any path.
     
  8. Oct 18, 2009 #7
    But this is a nonconservative force, where the work done DOES differ depending on the path taken. In fact, the question statements after the problem that the work done will differ, and the answers for the work done for each given path are different.
     
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