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Why is this not work?

  1. Mar 16, 2014 #1
    Two Scenarios:

    1. I have to push a cart. I apply a constant force of 10 newtons in the horizontal direction. The cart moves for 10 meters. I have done 100 Nm or Joules of work.

    2. I push the cart but I don't follow it and apply a constant force. It was just one moment of force, let's say 100 N. The cart moves forward, then friction eventually stops it after 10 meters in the direction that I pushed. Have I done 1000 Nm or Joules of work?

    I don't think I have, but can someone explain to me the difference?
     
  2. jcsd
  3. Mar 16, 2014 #2
    In (2) you accelerate the cart from rest to some velocity so you do some work. It may be over a short distance. But the force may be much larger than in (1) if you want it to move against friction over the 10 m.

    It's not clear what conditions you assume for (1). Is the force applied just equal to the friction, so that you have uniform motion?
     
  4. Mar 16, 2014 #3
    For #1 yes, I was assuming uniform motion.

    I guess I'm anticipating confusing questions I could be asked in the introductory physics high school course I'm teaching.

    Work = Force X Distance is not too hard of a concept. However, when I start to think more about the force it gets confusing to me.

    Is there a distinction between a constant force and a sudden force?

    Maybe I can ask again in a slightly different way.

    #1. Constant force of 10 N moves the object 10 m (with uniform motion) = 100 J of Work
    #2. Sudden push of 100 N moves the object 1 m (accelerates forward then friction brings it to rest) = 100 J of Work (?)

    Did I do the same amount of work?
     
  5. Mar 16, 2014 #4
    I saw a previous thread in which someone said as soon as you stop applying the force then work is zero. Not sure if that meant that the object had to stop moving or not. The physicsclassroom.com page said that a person throwing a shot-put is an example of work, which confuses me a bit more.
     
  6. Mar 16, 2014 #5

    Doc Al

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    Staff: Mentor

    This one's clear enough.

    Here you've done some amount of work, yet to be found. The cart moves while you push it, allowing you to do some work on it. Once you stop pushing, you no longer perform any work on it, regardless of how far it moves.
     
  7. Mar 16, 2014 #6

    Doc Al

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    Staff: Mentor

    Right.

    No, the object keeps moving until other forces bring it to rest. But that's got nothing to do with the work you did when pushing it. If you gave it a shove of 1000 N for 1 cm and it kept rolling for 100 m, you did 1000 x 0.01 = 10 J of work, not 1000 x 100 = 100,000 J.

    The person throwing a shotput exerts a force through a distance, doing work and giving the shotput energy. Sounds OK to me.
     
  8. Mar 16, 2014 #7
    Ah ok, that does sound more clear.

    So a person throwing a shot-put, or whatever object really, is winding up and exerting force on an object over the distance that they are winding up. That force X distance that they are winding up is the WORK. It doesn't matter if that shot-put goes 1000 meters horizontally, the amount of work that the thrower did is the same.

    If that is right, then I think I get it.
     
  9. Mar 16, 2014 #8

    Doc Al

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    Sounds good. Realize, of course, that the force exerted by the thrower is not constant as the object moves. But the basic concept is correct.
     
  10. Mar 16, 2014 #9
    I think I understand that. Variable force over a distance. It becomes a little more complicated to figure out the exact number for work. I'm going to avoid those situations for my students.

    Thanks for the help!
     
  11. Mar 16, 2014 #10
    Then the work done is the same in both cases.
    In the first case you had a force equal to the friction acting over 10 m. So the work is W1=Ff*10m where Ff is the friction force (equal to the pushing force).
    In the second case you accelerated the cart to such a speed that he kinetic energy acquired will be enough to move over the 10 m with friction.
    So the work done by your hand
    W2=KE (for the duration of the pushing force)
    and then
    KE=Ff*10m (for slowing down).
    You can see that W1=W2.

    The work-energy theorem is to analyse the second case.
     
  12. Mar 16, 2014 #11
    Unless it is a calculus based class, that would be a good idea. It does make sense to point out that there are techniques for dealing with such matters if the force can be written as a function of the distance. It might confuse some students, but I bet it will help some of them realize that the rules they are learning are only a special case which might make it easier to reconcile with their real world experience... or it won't. I teach calc and my "interesting" digressions rarely interest the kids.
     
  13. Mar 18, 2014 #12
    You only count the force and distance when the cart is in contact with your body. Once the cart separates from your body, your force on the cart is 0. If you push the cart and let go, then you only applied force over a short distance at the beginning.

    In your example, you pushed the cart with 100N over a distance of about 1m, then let go, and it went an additional 9m before stopping.
     
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