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What is the force work done by force F along OAC, OBC, OC

  1. Dec 13, 2004 #1
    B ------C (5,5)
    ..|...../|
    ..|.../..|
    O|/___|A

    A force acting on a particle moving in the xy plane is given by F=(2yi + x^2j) N, where x and y are in meters. THe particle moves from the origin to a position having coordinates x=5.00m and y=5.00m.

    What is the force work done by force F along OAC, OBC, OC.

    The answer books shows different answer. Woudln't the work just be mgh for all of them? Obvioulsy not, how would I solve this problem?
     
    Last edited: Dec 13, 2004
  2. jcsd
  3. Dec 13, 2004 #2

    Pyrrhus

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    By the definition of Work

    [tex] W = \int \vec{F } \cdot d \vec{r} [/tex]
     
  4. Dec 13, 2004 #3
    the integral of F=(2yi + x^2j) is

    F=y^2+.3x^3

    so W=(y^2+.3x^3)

    however, (x,y) is always 5,5, there isnt a change in work
     
  5. Dec 13, 2004 #4

    Pyrrhus

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    No, that's not true, for the first case you must evaluate from O to B, then from B to C and add both works for the OBC path.
     
  6. Dec 13, 2004 #5
    that doesnt seem to get me any new answers.

    O-->(0,5)B-->(5,0)C is W=(0^2+.3(5)^3)+(5^2+.3(0)^3)

    which equals (5^2+.3(5)^3)=66.6

    how is it possible to get different numbers?
     
  7. Dec 13, 2004 #6

    Pyrrhus

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    Because that force is not conservative, it depends on the trajectory.

    This means that the work done by the particle on a closed circuit or on a path from 1 to 2 and then 2 to 1 depends on the trajectory is not equal to 0, thus the force is not conservative.

    The following condition is not met:

    [tex] \oint \vec{F} \cdot d \vec{s} = 0 [/tex]
     
  8. Dec 13, 2004 #7
    I dont understand how the force can be different. Do you have an example?
     
  9. Dec 13, 2004 #8

    Pyrrhus

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    What do you mean not getting different answers???

    Probably your limits are wrong.

    For example for the OBC Path

    You first find the work from OB which means to Integrate from 0 to 5 with a differential dy

    [tex] W_{ob} = \int_{0}^{5} (2y \hat{i} + x^2 \hat{j}) \cdot dy \hat_{j} = \int_{0}^{5} x^2 dy = 0 [/tex]

    because x = 0 along this path

    Now BC

    [tex] W_{bc} = \int_{0}^{5} (2y \hat{i} + x^2 \hat{j}) \cdot dx \hat_{i} = \int_{0}^{5} 2y dx = 50 [/tex]

    because y = 5

    The work along OBC is 50 Joules.
     
    Last edited: Dec 13, 2004
  10. Dec 13, 2004 #9
    why is OB=0? there is a change in height.
     
  11. Dec 13, 2004 #10

    Pyrrhus

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    Because x = 0 and is considered a constant in the integral, what is considered a variable is what the differential tell us.

    [tex] W_{ob} = x^2 \int_{0}^{5} dy [/tex]

    Also Remember:

    [tex] \hat{i} \cdot \hat{j} = \hat{i} \cdot \hat{k} = \hat{j} \cdot \hat{k} = 0 [/tex]
     
    Last edited: Dec 13, 2004
  12. Dec 13, 2004 #11
    for:
    [tex] W_{ob} = \int_{0}^{5} (2y \hat{i} + x^2 \hat{j}) dy \hat_{j} = \int_{0}^{5} x^2 dy = 0 [/tex]

    why was the y value disregarded?

    for:
    [tex] W_{bc} = \int_{0}^{5} (2y \hat{i} + x^2 \hat{j}) dx \hat_{i} = \int_{0}^{5} 2y dx = 50 [/tex]

    how does:
    [tex]\int_{0}^{5} 2y dx = 50 [/tex]
    when
    [tex]\int_{0}^{5} 2y dx =y^2 = 25 [/tex]
     
  13. Dec 13, 2004 #12

    Pyrrhus

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    It was disregarded because of

    [tex] \hat{i} \cdot \hat{j} = \hat{i} \cdot \hat{k} = \hat{j} \cdot \hat{k} = 0 [/tex]

    and read above
     
  14. Dec 13, 2004 #13
    I'm sorry but I have no idea what that means

    Is there another way to approach this without integrals?
     
  15. Dec 13, 2004 #14

    Pyrrhus

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    Not that i know of, urban.
     
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