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Could someone explain why work is force times distance?

  1. Sep 25, 2012 #1
    I was wondering if someone could help me out with understanding the intuition behind multiplied units.

    For example, with speed, the units are divided (distance/time, m/s, etc.). The unit means how many meters you go in one second, which is the value received after dividing the distance by the time. But with work, it's force times distance (and with force, it's kg times acceleration). I'm trying to figure out what that represents, but I'm stumped. I was wondering if someone might be able to help me with this.

    Thank you
  2. jcsd
  3. Sep 25, 2012 #2
    Well, its generally a definition.

    We make up some magical quantity we call 'work', and define it (for seemingly no reason) to be force x distance. Then it turns out that this 'magical quantity' is really useful (like being equal to the change in kinetic energy), so we keep using it.
  4. Sep 25, 2012 #3
    So it's just a derivation of other principles? I thought maybe there was an intuition behind it
  5. Sep 25, 2012 #4


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    I think I remember reading that the concept of "work" as force times distance comes from the physics of simple machines such as levers, where it was observed that the force times distance is the same on the two ends of the lever. You can get a larger force (over a shorter distance) on one end of the lever by exerting a smaller force (over a longer distance) on the other side, but the amount of work is the same on both ends.
  6. Sep 25, 2012 #5
    Alright then. Anyone know how to mark a thread as solved?

  7. Sep 26, 2012 #6
    When the work consists of moving bags of cement upstairs, it's obvious that you will do twice the work, if you do 2 staircases instead of one staircase, and it's also obvious you do twice the work, if you carry 2 bags instead of 1.
  8. Sep 26, 2012 #7
    If I just hold the bag of cement in my arms, my arms get tired so presumably they are doing some kind of work? If that is so, then if a place the bag of cement on a table to give my arms a rest, is the table then doing work?
  9. Sep 26, 2012 #8


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    Neither you nor the table are doing work on the bag of cement - which is the relevant Force times Distance involved. However, your muscle fibres are doing work on bits inside your arm as they switch on for a while, move a bit, then relax. There is no corresponding movment inside the table. That's the difference.
  10. Sep 26, 2012 #9


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    Regarding the original question. Perhaps start with..

    Power = force * velocity

    To me that's intuitive. The faster you want to move a car the more power you need.

    power = dE/dt
    Velocity = ds/dt


    dE/dt = force * ds/dt

    Integrate to give

    dE = force * ds


    Work = force * distance.
  11. Sep 26, 2012 #10
    But on the surface of the table, aren't the individual atoms all jiggling around - and now and then push against an atom on the surface of the bag of cement and make that move against the force of gravity? So why aren't those atoms performing work in the same that the atoms in my muscle fibres are performing work against gravity?
  12. Sep 26, 2012 #11
    It is best not to talk about atoms and such in classical mechanics.
    Just think of matter as made of differential elements that still retain macroscopic properties.
    Of course, this it not true but it is enough for classical mechanics.
    You should look into continuum mechanics for non-particle interactions, (by particle I am not referring to atoms or electrons, simply point masses).

    In the case of the cement bag being on a table, no work is done since the bag is not displaced.
    Work is useful because the work done on an object is equal to the difference between it's final and initial kinetic energies. In a gravitational field, the work done is also equal to the initial gravitational potential energy minus the final gravitational energy (the opposite of the former).
    These facts lead us to the conservation of energy.
    Back to the bag of cement, since the potential energy of the cement does not change (it is at the same position), the difference is 0 and hence no work is done.

    If you are wondering why we shouldn't talk about atoms in classical mechanics, it is because newtons laws are not valid on that scale.
    Last edited: Sep 26, 2012
  13. Sep 26, 2012 #12


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    Did you ever see the Sourcerer's Apprentice in Disney's Fantasia? You asked a simple question and got a satisfactory answer but this is PF. The thread just won't stop. AAArrrgh.
  14. Sep 26, 2012 #13
    Force and distance are vectors. There are two mathematical operations we can do multiplying vectors, the dot product and the cross product. The dot product is a scalar, and this is work. The cross product is torque. The answer to your question is that we simply look at the two operations we can do with vectors.
  15. Sep 26, 2012 #14


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    Ten men working for two hours gives you 20 man-hours. That amount of effort can often be re-configurable so that four men could work for five hours and get the same result. That is an example where multiplying units yields a useful quantity - man hours.
    Likewise, with levers, you use Moments (a force times a distance) in the calculations. Does that help with "intuition"?
  16. Sep 26, 2012 #15
    Interesting comment. I like to view it as an example of applying mathematical principles to physics, multiplying vectors.
  17. Sep 26, 2012 #16


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    It's just another 'level' of approach. I could suggest that not everyone would involve vectors in their immediate 'intuitive' approach to a topic. They'd already be half way there if vectors were not a problem.
  18. Sep 26, 2012 #17
    We must teach the math behind the physics. Such as vectors.
  19. Sep 26, 2012 #18


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  20. Sep 26, 2012 #19
    Why eventually? Math is a collection of arbitrary consistent statements. Physics takes the logic of math and compares it with observation and experiment. This is a physical theory.
  21. Sep 26, 2012 #20
    That really helped. Thanks!

    Do these forums have reputation system (such as the kudos in dream.in.code)?
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