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When is work done on an object?

  1. May 4, 2013 #1
    Hi, this is my first post here.
    I have a degree in physics and i do some maths and physics tutoring.

    Recently I realised that I have some holes in my understanding of work, based on a question I was doing with a student (this is not the question I'm asking here though).

    I'll use a really simple situation that highlights my problem.

    A block is lifted up, using 1000 Joules of energy. How much work is done on the block?

    The way I would have thought of this previously is, 1000 J of mechanical work is done on the block. It gains 1000 J of gravitational potential energy. Therefore energy is conserved for the block.

    But the correct answer is NO work is done on the block. +1000 J of mechanical work is done on it, and -1000 J of gravitational work is done on it for a net work of 0 J.

    Initially, this seemed odd to me and made me think, if no work is done on the block, how can it be 1000 J better off in terms of total energy? Then I thought that although "the block" is better off by 1000 J, the source of the 1000 J of mechanical energy is 1000 J worse off, so everything still balances, and there are really 4 flows of energy when you count everything.

    BUT the correct textbook answer is to count only a particular two of those flows. The textbook answer that I read is that a change in GPE does not count in terms of work done ON something, because GPE is a property of the whole system, not just "the block", because there is another mass (in this case, the Earth) that is required for GPE to exist.

    The book only seems to count kinetic energy as belonging to the object itself. But that made me think, doesn't KE also require another object, or at least another reference frame, external to the object you are considering, for there to be a v in 1/2mv[itex]^{2}[/itex], the v must be measured with reference to something else. So the idea of "only counting KE" as work done ON something seems really arbitrary to me now.

    E.g. if the original situation is changed so that now the mechanical force is applied sideways instead of up (and everything is frictionless and lossless etc), and now 1000 J of work is done ON the block, legitimately, and the block has gained 1000 J of KE.

    What I am asking here is for any thoughts that might help clear up my understanding of this topic!
    Or, more specifically, some general rules for answering the question of how much work is done ON an object - which works count as done on the object, and which do not? Also, is there anything fundamental and intuitive to the answer to my question (which works count as done ON the object and which do not), or is it just a matter of convention and what is the accepted way that we all agree to define "on the object"?

    I think that the proper answer is going to be something in terms of identifying what forces are acting on the object. i.e. in the first situation, the mechanical lifting force, and gravity are the two forces acting ON the block. In the second (horizontal) situation, only one force is acting, the mechanical pushing force. But something still seems like its missing in my understanding.....
  2. jcsd
  3. May 5, 2013 #2


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    Ok, so we can solve this problem by being very strict in our definitions of work. First, let's make sure that we have a system; in this case, it's the block. Now, work is only defined for external forces. The person lifting the block is external to the system (the block), and the earth (gravity) is external to the block, so they both can do work, as your answer says.

    Now, let's be very strict in our definition of potential energy. Potential energy can only be defined for internal forces. If we want gravitational potential energy, the gravitational force between the earth and the block better be internal to our system (which we said was the block). If we wanted it to be internal, the earth and the block would both have to be in the system. So gravitational potential energy is actually undefined in this scenario.

    Keep in mind, the work done by gravity (when the system is the block) and the change in gravitational potential energy (when the system is the block and the earth) are equal in magnitude and opposite in sign. We can see that no matter what we choose the system to be, the physical results should come out to be the same!

    Kinetic energy is a property of the system, just like everything else! If our system is the block, then (1/2)mv2 works, so long as we're using the mass of the block and the velocity of the block. And you're absolutely right that you need to remember a reference frame every time that you calculate kinetic energy (usually we choose a frame at rest with respect to the earth). Notice something important: kinetic energy is not an intrinsic value; it changes depending on what reference frame you're in!

    I think you realize: choose your system and stick with it! Remember that work is defined for external forces, and potential energy is defined for internal forces. Yes, this is convention, but it is also the only way to formulate the theory consistently!
  4. May 5, 2013 #3
    I think the problem is the way the problem is phrased. It doesn't ask how much work a person must do to life the block. A person (or other device) must do an amount of work given my W = mgh where h is the distance that the block is lifted. The gravitational field does the opposite amount of work since the force exerted on it is in the opposite direction of the displacement. So the total work done is zero since the work done by person is the negative work done by the field leaving the total work done as zero. If there was a net amout of work done then the kinetic energy of the block would change.

    So if you wanted to know how much work that you must do to life the block then the answer is mgh,where hs is the value of the displacement and is positive in this case, and its positive since the force is in the same direction of the displacement,

    If you wanted to know how much work that the gravitational field must do to lift the block then since the force is in the direction opposite to the displacement then the work is -mgh. This time the force is in the direction opposite to the displacement so the work is negative.

    If there was a net amount of work done then there would be a change in kinetic energy.
  5. May 5, 2013 #4
    Thanks for your replies.

    I think I'm starting to feel like I get this. I did my degree in the 90s, and recently I've been tutoring -- mainly maths, and some physics, almost all senior high school level, and now I'm moving up to 1st year undergrad physics. I was really surprised that something from 1st year has taken me as much effort as this to get my head around.... (usually I can spend a few minutes with the textbook and it all comes back). But then it was a long time ago that I did my degree, and I've forgotten a lot of stuff.

    This is very helpful, I don't remember this from my degree. I think it basically fills the gap in what I was confused about. (The biggest part of it anyway). I think I was heading towards this in the final paragraph of my original post, which I wrote after the first part of it and hadn't thought of it before that... that was kind of a light bulb moment.

    Is this true always, even when other types of energy are involved (eg electromagnetic, chemical potential energy, or any other type of energy at all)? Or is this only true of mechanical work as opposed to work in thermodynamics which, as I understand it, is a more general use of the term work? (Again, pre-empting the answer to my own question, I'm thinking that its only mechanical work that this applies to, i.e. work of the type W=Fs (or [itex]\int[/itex]F[itex]\cdot[/itex]ds), and the change in KE is based on the fact that applying a force F through a distance s will always produce an acceleration and therefore a change in velocity and therefore in KE?)

    Also, when quoting, how do I get it to credit the person the quote is from?

    The original question that I was doing with the student was slightly more complex and it had a change in KE also. A cyclist was riding up a hill. The question gave the mass, the initial and final speed and the height of the hill. Part (a) was how much work was done on the rider and bike, and the answer (which confused me) was just the change in KE only. Part (b) was how much work was done by the rider, and the answer was as I expected. But I was expecting part (a) to be the same as (b).
  6. May 5, 2013 #5


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    This will be true for other types of energy, too! Consider a system of a proton and an electron, and you move them away from each other. Since they're both in the system, their Coulombic force is internal to the system, so we can say that their Coulombic potential energy increases, and it increases exactly with how much work you put in.

    Also, imagine a bomb in a controlled setting in a rocket ship. If it blows up, there is all sorts of chemical energy released, and the ship will move as a result. The chemical energy stored in the bomb is converted to work done moving the ship.

    For quoting, just push the little "Quote" button in the bottom-right corner of the post you'd like to quote. It should come up with the whole entry surrounded by the opening and closing quote commands. The top command should refer to the user and post. Just delete as much of the quote inside the commands as you want, and recopy the commands if you need to quote multiple parts.

    I agree with Popper, these questions are oddly worded. For part (a), we ought to choose the system to be the rider and the bike. If we remember that W = ΔE, and realize that the only kind of energy that is changing in that system is kinetic (remember, we cannot define gravitational potential energy without the earth in the system), we see that the net work done on the system is just equal to the change in KE.

    For part (b), I don't know what it wants the work done on. The earth? The bike? This is why we need to be very exact and systematic in looking at work and energy.
  7. May 5, 2013 #6
    Originally (before this question came up) I had thought of work quite loosely, in that anything that changed the net energy of a system meant that work had been done on it.

    After getting the wrong answer to (a) and not seeing how the ΔKE-only answer made sense, I started looking through 1st year undergrad textbooks, and I was seeing things like "Δ Gravitational PE doesnt count in the net work", and (even more restrictive), "only ΔKE counts as work".

    I've seen a lot of people posting on the internet (even in this forum) saying things like "KE has not changed therefore no work has been done", which only added to my growing doubts that i had something seriously wrong about my idea of what work is.

    Now I'm thinking that my original loose/general idea of work is still correct, providing that care is taken to identify what is part of "the system" and what is external to it, and adding the idea that work is defined for external forces, and potential energy is defined for internal forces.

    Is it pretty safe to say that that would be 100% correct? (lol i reread that sentence and it made me laugh.... like saying its correct 80% of the time, every time).

    I was thinking of Coulombic potential before also, (and that was confusing me even more) since its law is so similar to Newton's gravitational law, and wondering whether, under this new, mysterious idea that some energy changes are not work, it would count. I.e. thinking that just by symmetry, if the original problem I gave of the block being raised against gravity wasn't work, then an identical problem with a charge being moved against an electric field should also not be work.

    I'm thinking now that the idea of being very careful about defining what is in the system and what isn't will cover all situations.

    Just to settle the Coulomb issue in my mind, if a proton and electron were moved apart slightly in an atom, then it would be correct to say work has been done on the atom, but not on either of the individual particles (counting them individually and without the other particle or the atom)?

    Thanks, is there any way to quote posts from more than one person in the same reply by this method? (or any way better than opening the thread page in multiple windows, and clicking the quote button for each person, and copying/pasting the QUOTE tag with their ID number?

    The question was problem 6.73 from University Physics by Young, 12th Ed. For part (b) the specific wording was "how much work have you done with the force you apply to the pedals" which I took to mean how much work is done by the rider and the the bike (ie the same system as in part (a)).

    But I wasnt sure if it was ok to quote an actual question in this forum , since there is another forum for questions.... I put it here since I know the answer and I wanted to know more about work in general (and how to explain work to my students).
  8. May 5, 2013 #7
    P.S. when i was originally thinking about what is doing work on what, (i.e. the rider and the bike), it made me wonder whether something could do work on itself or not. I read about this for a while (using Google), and found several conflicting opinions, and I settled on the idea that a rigid body could not do work on itself, and only a non-rigid body could do work on itself. This makes sense to me based on the idea that W=Fs and there must be a displacement for there to be work, and if you are talking about a body doing work on itself, then the displacement s must be internal to the body itself (i.e. non rigid). I was happy with this idea, would you say it is correct?
  9. May 5, 2013 #8


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    Yes; as long as using Newton's second law is valid, this approach will work!

    Well, that depends on both your system and your reference frame (remember, reference frames matter here). If our system is the electron, and we're in a frame of reference at rest with respect to the proton, then we can say that work is done on the electron in moving it away! In another reference frame (e.g. at rest with respect to the electron), the electron is not displaced, and so in that reference frame, there is no work done on the electron. Keep in mind that when you choose a system and reference frame, you need to keep it consistent for both work and energy.

    Certainly! There is a small "M" button next to the "Quote" button. If you click that on multiple posts and then click "Quote" somewhere, it'll open the same as before but with multiple quotes!

    I have this text, and I looked over it, and I have to argue that a more appropriate phrasing would be "how much energy do you expend with the force you apply to the pedals?".

    In order to solve this one, we'll choose our system to be the bike and the rider (since we only know the combined mass). For the work-energy theorem, the work done is only by gravity, and the change in energy is both kinetic and "internal" (your muscles must lose some chemical or other energy in powering this process). We're left with:
    [tex] -mgΔh = \frac{1}{2}m(v_f^2 - v_i^2) + ΔE_{internal} [/tex]
    For the [itex] ΔE_{internal} [/itex], we get the same magnitude as the book, but we get a negative value (we should expect that energy to be lost!) If you want to call this "work done," that is probably valid in some conventions. I wouldn't call it that, because it'd be very hard to formulate consistently. Either way, our approach still works, we just have to change our question a little to make the conventions match.

    At a deep (beyond introductory physics) level, I might agree with that analysis. Remember, though, that work can only be done by external forces, and if the system is supplying the force on itself, then the force is internal! Work done "internally" to a system is usually just looked at in terms of energy; like the energy from a bomb being converted to kinetic energy as we discussed.
  10. May 6, 2013 #9
    I think I'm starting to get this now.

    Ive been doing some more 1st year undergrad level questions (this time from Halliday and Resnick 9th ed), and the ones from it that Ive seen so far that have this type of situation, HAVE counted work done against gravity (even with no change in KE) as work in the answers to the questions --- just as I would have done myself all along, before I ever saw the original question that I posted about that confused me.

    I had to think a bit before I saw the difference in the wording of these questions

    eg Halliday and Resnick ch 7 question 23. a box is being dragged up a slope by a force F through a vertical distance h, at constant speed. How much work is done on the box by F? The Halliday and Resnick worked solutions say that since there is no change in KE, the work done must be equal to the negative work done by gravity, and the answer is +mgh. (there are more numbers in the actual question, which i could quote if it would help, but this is the guts of it). This is exactly what I would have thought to do before.

    There was another question, which also gave the ΔGPE as part of the answer, and again in that question the wording asked how much work was done by the pulling force?

    The question from Young that I got confused with, asked what was the total work done on the bike and rider. Implying that they are including the negative work done by gravity plus the positive work done by the rider against gravity, which sum to zero, leaving only ΔK as the answer.

    Now that I feel that I understand this, the two questions from Halliday and Resnick that Ive discussed in this post seem like "real" questions in the sense that the answer is something that you might really want to know in a real life question. The question that initially confused me seems more like a trick question, designed to highlight this whole issue that I have raised, so it seems like more of a teaching question rather than one which reflects an actual problem that someone might want to solve.
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