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Work done by a conservative force

  1. Aug 22, 2015 #1

    Titan97

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    1. The problem statement, all variables and given/known data
    The potential energy ##U## of a particle of mass 1kg moving in x-y plane obeys the law ##U=3x+4y##. x and y are in meters. If the particle is at rest at (6,8) at time t=0, then find the work done by conservative force on the particle from initial position to the instant when it crosses the x-axis.

    2. Relevant equations
    ##\vec F=-\frac{\partial U}{\partial x}-\frac{\partial U}{\partial y}-\frac{\partial U}{\partial z}##
    (This equation is not taken from any book. I thought the relation between F and U was just like the relation between electric field and potential)

    3. The attempt at a solution

    Using the equation: ##F=-3\hat i-4\hat j##
    Since no other forces are acting, the particle will move in the direction of acceleration. I also have to find the x-coordinate when it crosses x-axis. Acceleration is at an angle ##\tan^{-1}\big(\frac{4}{3}\big)## with the horizontal towards the 3rd quadrant. Hence the particle moves along the line ##y=\frac{4}{3}(x-6)+8##.
    So the x intercept is 0. Hence the total distance moved by the particle is 10m. And work done is 50J.
    Is this correct?
     
  2. jcsd
  3. Aug 22, 2015 #2

    BvU

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    It is. A lot of work done by you. The work done by the field can also be compared to the difference U(6,8) - U(0,0). Could that be a concidence or is there more to that ? :rolleyes:
     
  4. Aug 22, 2015 #3

    Titan97

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    Work done by conservative force is path independent. But I still need to do "work" to find the final position :smile:. Or is there another way to solve the problem without having to find the final position?
     
  5. Aug 22, 2015 #4

    BvU

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    Force always points at (0,0) !
     
  6. Aug 22, 2015 #5

    Titan97

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    Why should it always point at (0,0)? Is it because the force is conservative?
     
  7. Aug 22, 2015 #6

    BvU

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    No, but starting at (6,8) there is no component other than in the direction of (0,0)
     
  8. Aug 22, 2015 #7

    Titan97

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    I dont understand that. Is it because the minimum magnitude of potential energy is at (0,0)? Since a particle tends to reach minimum potential energy. If thats the case, then is it true for ##U=2x^2+1##? Since force points at (0,1).
     
  9. Aug 22, 2015 #8

    BvU

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    Potential energy is a scalar. It has a value. Vectors have a magnitude. Unfortunately daily language mixes them up.

    Lines of constant U are straight lines for ##
    U=3x+4y##. Constant U means no force component along that line. The force points perpendicular to those equipotential lines, so once on a line through the origin means following a straight path through the origin if starting from x > 0 and away from the origin when starting from x ##\le## 0.

    For ##
    U=x^2+1## lines of constant U are straight lines also. The force always points towards the y axis, not at 0,1 !
     
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