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Calculate the work done by a force

  1. Dec 13, 2005 #1
    I am asked to calculate the work done by a force as it moves around a path. The force is F = b(1-x^2/a^2)j. The path is a rectangle with coordinates at (0,0); (0,L); (a,L); (a,0). The force moves clockwise around the path beginning at the origin. A diagram is attached.

    I know work is the integral of F dot dr.
    So for the first path I should have the the force F=b(1-x^2/a^2)j dotted with Lj (the path from the origin to point (0,L)). The integral is thus bL (1 - x^2/a^2) dy with limits from y=0 to y=L. Is this the right approach? If not, can someone please point me in the right direction??

    Attached Files:

  2. jcsd
  3. Dec 13, 2005 #2

    Physics Monkey

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    You have an extra factor of L that you don't need (look at the units). The work is the integral of [tex] \vec{F}\cdot d\vec{r} [/tex] along the path. For example, the first segment has [tex] d\vec{r} = dy \hat{j} [/tex]. You have to figure out what [tex] d\vec{r} [/tex] is for each of the four segments.
  4. Dec 14, 2005 #3
    So, for the first segment, dr = dyj. For the second segment, dr = dx i.
    For the third segment, dr = -dyj. For the fourth segment, dr = -dx i. Is this correct? Are the limits on my integration correct as well?

    Also, should the answer be 0 (closed path, conservative force...not sure if the force is conservative though)?
  5. Dec 14, 2005 #4


    Staff: Mentor

    If the answer is zero then the force is conservative, but not all forces are conservative so you can't use that as a check here. (Your dr vectors are correct).

  6. Dec 15, 2005 #5
    I get an answer of 2bL (1- x^2/a^2). This does not seem correct to me, since it contains an x^2 term? Is this right? Is there a substitution I can make for x? x=a or x=L, for instance? This problem is driving me crazy...any help greatly appreciated!
  7. Dec 15, 2005 #6


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    You can eliminate two of the legs from your problem since the force is in the [tex]\hat{j}[/tex] direction.

    In segment 1, x=0. In segment 3, x=a.
    Last edited: Dec 15, 2005
  8. Dec 15, 2005 #7
    Yes...I have the forces for the dx direction to be zero. I'm still doing something wrong though?
  9. Dec 15, 2005 #8


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    Did you substitute in the values for x that I just edited into my last post?
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