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Work Done

  1. Sep 17, 2005 #1
    I have problem on this am not sure with my solution.

    A force of x-y plane is given by F = A(10ai+3xj),where A and a are constants.F is in Newtons and x is in meters.Suppose that the force act on a particle as it moves from position x=4m,y=1m to a final position x=4m,y=4m.Show that this force is not conservative by computing the work done by the force for at least two different paths.

    I have tried to draw the xy axis diagram and getting the rectangular with x=4 and y=3,do I pluged it in scalar product of W.D=F.X AND F.Y respectively? am I right?
  2. jcsd
  3. Sep 18, 2005 #2
    You need two different paths, say r1 and r2, which move between (4m, 1m) and (4m, 4m). Then compute the work done by the force on each of these two paths, and compare the results. If the force is conservative, you will get the same answer no matter how your particle travels between these two points.

  4. Sep 18, 2005 #3
    Break F into [itex]F_x[/itex] and [itex]F_y[/itex] and use

    [itex]W_x= \int F_x dx[/itex] and similarily for W_y , since work is a scalar add them , now try to calculate the swork through a different path . and see if both turn out to be the same

  5. Sep 18, 2005 #4
    Thanx u all
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