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Calculate for each field the work done in moving a particle

  1. Feb 5, 2004 #1
    Problem:

    Which, if either, of the two force fields F1=-yi+xj+zk, F2=yi+xj+zk is conservative? Calculate for each field the work done in moving a particle counterclockwise around the circle x=cos(t), y=sin(t) in the (x,y) plane.


    Answer:

    I got the part in telling if the force fields are conservative. F1 is not conservative because d/dx(Fy)-d/dy(Fx) is 2. F2 is conservative because everything is equal to zero.

    How do I calculate the work done in each field? I don't get it.
     
  2. jcsd
  3. Feb 6, 2004 #2

    HallsofIvy

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    If a force field is conservative, then work done is the difference in potential energy at the two end points. In particular, the work done in moving around a closed path is 0. (The work done in lifting a mass against gravity can be regained by dropping it back down.)

    If a force field is not conservative, then you will have to integrate the dot product of the force with ds around the path.

    In this case, x= cos(t), y= sin(t), z= 0, so ds= dxi+ dyj+ dzk=
    -sin(t)i+ cos(t)j+ 0k. F . ds= (-yi+xj+zk).(-sin(t)i+ cos(t)j+ 0k)
    = (-sin(t)i+ cos(t)j+0k).(-sin(t)i+ cos(t)j+ 0k)= sin22(t)+ cos2(t)= 1. The work done is just the integral of the constant 1 around the circle and so is just the circumference of the circle.

    If you did integrate F2 in the same way, F2.ds= (yi+xj+zk).(-sin(t)i+ cos(t)j+ 0k)= (sin(t)i+cos(t)j).(-sin(t)i+ cos(t)j+ 0k)= cos2(t)- sin2(t)= cos(2t) and the integral of that around the circle (t going from 0 to 2π) is 0.
     
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