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Work done in partial circle

  1. May 28, 2012 #1

    I understand that work done in a conservative field when a closed loop is followed is zero.
    The answer to this question I am sure is B. How do I explain to my colllegue that it is not zero when he thinks that the speed is constant and there is no friction and the force is at 90 degrees to the velocity vector.
  2. jcsd
  3. May 28, 2012 #2
    Hello Dave, perhaps if you were to explain what you mean some sort of help might be forthcoming.

    What is B?
  4. May 28, 2012 #3
    I am afraid I was unable to post the image I uploaded,
    The image is hosted at http:// imgur.com/kBTVm
    spaces need removing
    Any help would be greatly appreciated
  5. May 28, 2012 #4
    Sorry but I get that page as unavailable.

    In any case reference to online images etc is poor practice because that webpage may well have disappeared when someone comes to view the thread in the future.

    So just explain as best you can.

    Note you can go a long way writing formulae with the sub abd superscript Xs[/SUB2 and X2 in the bar above and the selectable symbols on the right.
  6. May 28, 2012 #5
    OK I have managed to see your webpage - (spotted the space a bit late).

    Why do you think your friend is wrong?

    Edit it might help to state your definition of work.
  7. May 28, 2012 #6
    A particle of mass m moves horizontally at a constant speed v along the arc of a circle from p1 to p2 (the diagram indicated a quarter of a circle) under the action of a force. What work is done on the particle by the force during this displacement?
    a) Zero
    b) ((pi)mv^2)/2
    B being from force (mv^2)/r and displacement (2(pi)r)/4

    Thanks for the tips, I will ignore my sons helpful ideas of using imgur, I couldn't even pronounce it anyway.
  8. May 28, 2012 #7
    I am happy that the work is zero in a closed loop if the force in question is conservative.
    He is saying that no work is done because the force is alway at 90 degrees to the velocity.
    He also is saying that the kinetic energy is constant and that there is no friction so nothing can convert another form of energy into the work
  9. May 28, 2012 #8
    Last time I walked around an equipotential path around a mountain I still had to have a lie down
  10. May 28, 2012 #9
    OK we know the question.

    The issue is that you are having trouble arriving at the correct answer and don't agree with your friend's explanation?

    What is your definition of work?

    Consider a block of ice being slid across a flat level tabletop at constant velocity.
    The block of ice is subject to the force of gravity throughout but no work is done. Can you see why from the basic definition of work?
  11. May 28, 2012 #10
    So the block of ice is sliding on ice at constant speed implying no horixontal forces. Gravity pulls the block downwards and is always acting at 90 degrees to the velocity so the dot product of the two vectors is always zero

    Thanks for the help
  12. May 28, 2012 #11
    Dot product, energy, conservative force, closed loop...

    Let's keep it basic.

    KISS (Keep it simple stupid)

    Work is done when a force moves its point of application.

    The work done equals the product of the magnitude of the force and the distance moved in the direction of the force.

    First the ice block.

    The ice block moves neither up nor down in its travels. So the distance moved against/with gravity equals zero.

    Anything times zero is still zero.
    In particular the work done is zero from the definition.

    Now your example is more tricky since your particle is moving in a circle.
    The only possible force that can act is directed along a radius from the particle towards the centre. It is this force that causes the particle to deflect from a straight line.

    However this force must be constantly changing direction to be always radial.

    Because the path is a circle the radius does not alter. That distance the particle moves against/with the direction of the force is zero.

    Again zero times anything is zero so the work is zero.

    This type of motion is called motion under a central force and is like whirling a stone around your head on a string.

    BTW you should really post this in the homwork section. I am not a moderator, but you might find a moderator will move it.
  13. May 28, 2012 #12
    Thanks Studiot, I like the concept of keeping things simple. I can see why the answer really is A then. This is not homework, unfortunately it is an old exam question and the exam board said the answer was B not A. It caused a bit of a debate amongst a few of us.
    Thanks for your help, alas I am off to do work against gravity and climb the stairs at 180 degrees to the force.
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