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No work done (apparent paradox) What am I missing?

  1. Dec 9, 2007 #1
    No "work" done (apparent paradox) What am I missing?

    I have a simple question about work. Work is force times distance, in the direction that the force is applied. So when I carry a box, at constant velocity, at constant height, we draw a free-body diagram like that shown below. Since the force applied is perpendicular to the motion, we say that the total work done on the box is zero. (See figure.)

    [link fixed]

    Here is my problem: Why are we only looking at one force (vertical)? Isn't there a horizontal friction force? Why is this force not considered?

    I know that we don't change the box's velocity; we don't change its PE or KE. So if no change in energy occurs, no work is done. Fine. But why don't any textbooks mention the existence of a sideways friction force (without which the box would not move?) I get the feeling that the textbooks are making an over-simplification, and leaving out essential need to know information.

    I have noticed some very bad problems shown in otherwise good books. Books say that in a given problem, a body in motion across a tabletop requires zero force to stay in motion. In book after book, test after test, I see this, without the provisio being given that this is only true when no friction exists.

    When friction exists a force must be applied continuously (to counter the friction force, so the net force is zero.) Usually the textbooks get it correct in one paragraph, but then explain it wrong everywhere else (and also in most test questions.)

    Is it no wonder that so many of our students get some answers right, but revert to Aristotelian physics when they try to understand a situation?

    What am I missing in the original question about work? How is it that we do zero work, even though we are applying a horizontal force (which is in the direction of motion)

    Last edited by a moderator: Dec 9, 2007
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  3. Dec 9, 2007 #2

    Doc Al

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    Please check your link--I can't seem to view it. But it's certainly true that to get the box moving at all requires some work to be done on it and a force exerted on it. But once you get it moving, no additional work is done on it. What friction force are you thinking of? (Of course, we neglect air resistance.)

    That's pretty bad. Any time you slide over a surface, the existence or non-existence of friction should be mentioned. (What physics book ignores this?) Of course the net force is zero.

    But I feel your pain: Basic physics errors in textbooks are a pet peeve of mine and I'm always on the look out for them.

    You only apply a horizontal force to accelerate the box to whatever speed you are carrying it at, which can be made arbitrarily small (if you've got the time). Once it's moving, the only horizontal force needed is to counter air resistence.
  4. Dec 9, 2007 #3
    No "work" done - What am I missing? (Possible way to phrase the answer.)

    Ok, that's what I thought.

    * When you pick up the box from the ground, you do work.

    * Standing still with the box in your arms, you do no work.

    * When you start to move, you are accelerating, and until you reach a maximum speed, you are doing work on the box.

    * Once you reach top speed, and continute at this same speed, can I say that no net work is done on the box?

    * The box, in practical terms, feels two retarding forces against its direction of motion:
    (a) air friction against the box
    (b) it is pulled back (inadvertently) by me, due to air friction (slight) and friction of my feet against the ground (significant)

    * However, I constantly add a forward horizontal force to overcome these frictional forces. (This is the sideways friction force I was mentioning - the horizontal force of my hands against the box, pushing the box forwards, to counteract the various drag forces.)

    * The net force on the box is thus zero, the acceleration is zero, and therefore no net work is done on the box.

    Would this sequence of events be correct? If so, I think this is a better way to explain this to students, because they violently rebel against the idea that "no work is done", when they plainly feel effort as they do this experiment themselves.

  5. Dec 9, 2007 #4

    Doc Al

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    Right, because the force you exert is in the same direction as the displacement. (Of course, the net work on the box may well be zero if the box is lifted at constant speed.)

    Right. A force with no displacement does no work. (Despite the effort it may require to support the box.)

    Right. Since the horizontal component of your force is in the direction of the displacement.

    Sounds good. If something is moving at constant velocity the net force, and thus the net work, is zero.

    I understand what you're trying to do, but I find it a bit confusing. Are you modeling yourself pulling and pushing the box? I think it's a bit distracting. I prefer to keep it simple: If you are able to keep the box moving at constant speed, then the horizontal force you are exerting is either zero (ignoring air resistance) or small (equal to air resistance). Personally, I wouldn't even bring up air resistance except to say we will ignore.


    But the reason they feel effort has little to do with the forward motion of box (and the details of walking) and more to do with biology. Unfortunately, as often happens, the commonly used word "work" has been co-opted by physics and given a specific technical meaning that has little to do with its everyday meaning. So I'm not sure how helpful such an analysis (as attempted above) will prove.

    A more direct approach would be to ask why does it require "effort" to just hold up a box? Or just push against a wall, for that matter. That's where a more in depth analysis of muscle fibers contracting and relaxing (doing work!) just in order to maintain tension may prove helpful. There is work going on, but that work is done internally, not externally against the box (or wall).
  6. Dec 30, 2007 #5
    How to phrase answers to question about work?

    Question: A waiter carries a tray full of meals across the room. Is work being done?

    My first response: We can’t answer the question yet, because the question is flawed!Questions about work only make sense when they ask: Does (something) do work on (some object)?

    Also, questions only make sense when the motion is precisely described. So let’s restate the question. In fact, let’s ask three different questions.

    A. Does the waiter do work on the tray, as he starts from rest and then reaches some final velocity? YES

    B. Does the waiter do work on the tray, as he slows down to a stop? YES

    C. Does the waiter do work on the tray, as he moves at constant velocity? NO. To be more precise, NO NET WORK is done by the waiter on the tray.

    But here is what I want to examine: “While moving at a constant velocity, isn’t friction between the waiter and the floor doing work on the waiter?“

    I want to say "Yes, the friction force between the waiter and the floor constantly retards the motion of the waiter (and thus also would retard the motion of the tray.) The friction force acts over the distance that the waiter is traveling;
    the friction force thus does work on the waiter."

    Is this correct to say? Can I say that the friction force effectively doing work on the waiter’s muscles?

    I also want to say "However, the waiter constantly uses his muscles to overcome this friction force, and continue moving forward at constant speed. The waiter thus does work that is equal to the work of the friction force, but in the opposite direction."

    Can I say that he does work on, or within, his own muscles?

    As you can see, I wish to explain something like: The work of A on B is cancelled out by the work of B on A, hence the NET work on the TRAY comes out to zero.

    But how specifically should I describe this? Is my description ok? I have the feeling that it is almost correct, but I just want to be as precise in my wording as possible. (I've seen too many questions that were worded, or answered vaguely, even in textbooks.)

  7. Dec 30, 2007 #6
    They feel it is work because they have to counter the force of gravity by using their muscle :P
  8. Dec 30, 2007 #7
    you see either i dont understand what you mean or this state ment is wrong

    work is not a vector quantity or in better words doesnt depend upon direction
    yet you see net energy gain by tray is 0 because work gained = work lost hence the tray doest not get any energy from it

    but you consider system of two protons (they repel each other)
    now one pro. do some work on other and other does on this one but work doesnt get cancelled out net work done in the process on the system is W1+W2

    the sign assigned to work tell wther work is done on the system or by the system and not
    bodies taking part .

    you do not get work converted into K.E or potential energy
    coz. .friction is a nonconservative force hence unable to store or release potential energy
    what it does is just dissipation...so when waiter moves with constant velocity he does work against air resistance

    when you stand holding something then you do any work but still can feel the effort coz. feeling is just a sensation it not refers physical fatigue due to work done but also the pressure and tension in the muscles
    while you have to tighten your muscles thus increasing tension which you can feel as pain
    quoting : physical truth does not always depend on what you feel or what you not , it just depend on what god feels
    Last edited: Dec 30, 2007
  9. Dec 31, 2007 #8

    Doc Al

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    Sounds good.

    Sounds good.

    If the tray is being moved at constant velocity, I'd say that the waiter does no work at all on the tray. His force on the tray? Vertical. Direction of motion? Horizontal. Work = 0.

    Any friction of the floor against the waiter's shoes does no work at all on the waiter. It's just static friction, there's no movement of the point of contact.

    In any case, this friction force is exerted on the waiter, not on the tray.

    It does require energy for us to walk, but that's due to the biomechanics of walking, not because of friction from the floor. (Of course we rely on that friction to push against.) We are constantly stopping and starting our legs, pushing and braking, in order to maintain a constant speed of our torso. So, yes, internally, within the muscles, we are doing work as we tense and relax various muscle fibers.

    Again, I'd say that if the waiter is able to keep the tray moving at constant velocity then he's doing no work at all on the tray (not just no net work).
  10. Dec 31, 2007 #9
    Negative work cancels out positive work.

    I think I see the problem here. Is anyone else considering negative work?

    On occasion, a force acts upon a moving object to hinder a displacement. Examples might include a car skidding to a stop on a roadway surface or a baseball runner sliding to a stop on the infield dirt. In such instances, the force acts in the direction opposite the objects motion in order to slow it down. The force doesn't cause the displacement but rather hinders it. These situations involve what is commonly called negative work. The negative of negative work refers to the numerical value which results when values of F, d and theta are substituted into the work equation. Since the force vector is directly opposite the displacement vector, theta is 180 degrees. The cosine(180 degrees) is -1 and so a negative value results for the amount of work done upon the object. ​

    Consider this problem. A 10-N force is applied to push a block across a frictional surface, at constant speed, for a displacement of 5.0 m to the right.

    There are four forces. Two of these forces do work. Even though the object moves at constant velocity, work is done on this object! However, the net work on the object adds up to zero.

    F(applied) and F(friction) do work. F(gravity) and F(normal) do not do work, since a vertical force cannot cause a horizontal displacement.

    The forces that do work can be calculated; they add up to zero.

    Wapp = (10 N) * (5 m) * cos (0 deg) = +50 Joules

    Wfrict = (10 N) * (5 m) * cos (180 deg) = -50 Joules

    This is what I am talking about with the waiter and the tray problem

    Cool. But how should I explain that a person, walking on a surface with friction, tends to comes to a stop. (Not just eventually, but feels a retarding force at every moment.) Should I say that walking people come to a stop due to:

    (a) static friction between feet and floor. (No work done here; the floor doesn't move.) and also

    (b) internal friction in the person (between bones and tendons, between muscle fibers), some of which produces internal work. (i.e. force of one body on another body component.)

    Ok, here is where I have some difficulty in how things are phrased. With friction, all objects come to a stop, due to the sum of all the retarding friction forces. So not only does it take energy for us to walk, we constantly have to apply an external force against the ground (and also, against the internal parts of our body) to stay in motion.

    So can I say that moving forward, at constant speed, involves two components:

    (A) Internal work, within the body, as you described, and

    (B) Force of feet against the floor. By Newton's 3rd law, the floor puts and equal and opposite force on us, propelling us forward.

    Now, when we put our feet against the floor, the floor doesn't move. We thus do no work against the floor.

    But the floor does put a force on us which moves us forward. Can't we say that this is producing work on the person. I am trying to carefully distinguish between:

    (a) work done by one object on some final object, and
    (b) the net work done on some final object.

    I have seen problems done where (a) shows some work being done, and (b) shows that the net work is zero. That is what my question is all about. The net work done on tray is zero. (Also, net work done on the waiter is zero.)

    This is what confuses me; are you considering the existence of negative work, or is negative work a formalism that you think is ill-advised?

  11. Dec 31, 2007 #10

    Doc Al

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    Realize that in this example (and in those quoted from the website):
    - There are two forces doing work on the object;
    - The friction force doing the negative work is kinetic friction, not static friction.

    But in your waiter/tray problem:
    - There is no force doing work on the tray;
    - The friction force that acts on the waiter is static friction, which does no work.

    The biomechanics of walking is non-trivial. It's not a simple matter of overcoming friction with the ground. Just the opposite--friction helps us walk. (Try walking on ice.)

    Any work being done is being done internally. If our walking were more efficient, it would require less energy. (Part of our gait can be modeled as an inefficient inverted pendulum motion.)

    It's certainly true that as we walk friction with the ground pushes our feet one way, then the other.

    Again, while friction does accelerate and deaccelerate the waiter, it's incorrect to say that it does any work on the waiter, negative or otherwise.

    I just don't think it applies to this situation.
  12. Dec 31, 2007 #11
    Got it, thanks much. I think I was getting myself in way over my head by trying to take these considerations into account. I imagined that there was much simpler way to model this, but now that I think about it, its very complicated.

    Last edited by a moderator: Dec 31, 2007
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