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Displacement and Work

  1. Jun 29, 2009 #1

    I am taking my first semester of physics this summer and am having trouble grasping what I imagine is supposed to be a pretty fundamental idea.

    I understand that displacement is a vector and is dependent only on the final and initial points, not the path taken. What I cannot resolve in my head is how this works in the "real world." That is, if a car drives in a circle of 2pir, it did zero displacement. If it drives a straight distance equal to 2pir, it has displacement. How does this work? Doesn't the engine burn fuel, the odometer advance, etc no matter if the car is going in a circle or a straight line? How can the two be different, and what is the physical difference in terms of work done?

    Sorry if this is a silly question.
  2. jcsd
  3. Jun 29, 2009 #2
    Not a silly question. If work were in fact force times displacement, yours would be a very good question.

    But it's not. Work is force across distance, not displacement.

    Your intuition was right.
    Your definition was wrong.

    Make sense now?
  4. Jun 29, 2009 #3
    Um, I'm afraid archosaur is incorrect. Given a CONSERVATIVE force (mathematically we say that it is a field such that grad x F=0, if you don't know what this means don't worry) the work done through a closed loop is indeed zero. Examples of conservative fields are gravity, electric attraction, etc. Now as for your car example, you are correct that you will lose energy when traversing the track (after all, it takes gas) so what gives? The reason is that friction and air restistance ARE NOT conservative forces so while traversing the loop you leak energy. In addition, if you did this experiment on a frictionless track in a vacuum you'd notice nothing would happen, the reason being that those are actually the only two forces at play. The car is not enticed to traverse the track by anything other then the turning of the wheels and the turning of the wheels only produces motion because they have friction with the ground and so on...
  5. Jun 29, 2009 #4
    If you rub a pencil eraser back and forth 100 times on one centimeter of paper and there is a kinetic frictional force of (for calculation's sake) 2 Newtons, the eraser will be back where it started, but you will have converted working energy into 2 joules of heat energy.
  6. Jun 29, 2009 #5
    Because friction is not a conservative force. Take a postively charged ball and put at point A and then take it on all sorts of crazy paths and loops and take it back to A. It'll have just as much energy as when it started.
  7. Jun 29, 2009 #6


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    Work actually is force times displacement, but only for infinitesimal movements. So, to take the car example, when you drive a car around in a circle you can't just note that the net displacement is zero and conclude that the work is zero. You have to take it a little bit at a time. So the work done in the first little bit of motion is the force exerted times the first small bit of displacement, and the work done in the next little bit of motion is the force times that next small bit of displacement, etc. etc. etc. all the way around the circle. If you know calculus, you'll recognize this as a non-mathematical description of
    [tex]W = \int \vec{F}\cdot\mathrm{d}\vec{s}[/tex]
    For conservative forces, it just happens that the result of adding up all those little bits of work depends only on the net displacement.
  8. Jun 30, 2009 #7
    for to be work performed, there should be non zero displacement
    But this doesnot means the energy is not spent
    Every energy spent cannot be able to do displacement
    For example - you push the wall ,but is not displaced , work is not done but your valuable energy is spent
    The difeerence is only that:wink:
  9. Jun 30, 2009 #8
    This is almost true, and brings up a good point:
    As in Starproj's initial example, we see that net displacement is not required for work to be performed (the car returns to its initial position). The energy that is lost in driving your car around a circle is precisely due to the work that must be done because of the nonconservative nature of the frictional force involved.
  10. Jun 30, 2009 #9
    There seems to be a lot of confusion in this thread. First off, work is a force times a displacement. A force doesn't have to be applied over a distance, you could also have a pressure applied across an expanding volume.

    Second, you need to distinguish between NET work and just work. In your car example the net work is indeed zero. Image throwing a ball up in a vacuum. You are applying a force along a displacement to the ball when you throw it up but on the way back down the ball is applying a force over a displacement to your hand when you catch it, equal to the force and displacement required to throw it up. So,

    work to through it up - work to catch it = 0 = net work

    Lastly, you need to understand the difference between energy and work. They are not the same thing. In your car example the net energy will be negative (depending on how your system is defined) but your net work will be zero.
  11. Jun 30, 2009 #10
    So, work is energy spent against a field force, such that energy is stored?

    aka, lifting a weight, or pulling back a rubber band?
  12. Jun 30, 2009 #11
    Oh wow there's a lot of confusion here. This should be pretty basic people! Work is a line integral of the form F dot dr integrated over a curve. For closed curves (i.e. ones that start where the end) it is easily shown that if F is conservative force (i.e. curvF =0) then the net work is zero. However, the car example IS NOT a conservative system since their is friction and air drag, this is why when the car returns to its initial position it has lost energy (you've used fuel).

    Net work and 'work' are identical. I don't really understand what difference you're trying to point out. If you look at the work done for only, say, the first quarter of the track rather then the complete loop, you're still looking at the NET work.

    Finally, work IS the amount of energy expended. That's exactly what it is. Energy is often defined simply as the capacity to do work. The examples mentioned of pushing on a wall expending energy but not performing work is incorrect (Feynman in his lectures on physics has a really good discussion of this confusion). The problem is in the way our skeletal muscle has evolved such that to apply a constant force with our muscles we must constantly create electric gradients in our muscles and discharge it. Therefore, to maintain a constant push force we constantly use energy (smooth muscle does not behave this way). This however, DOES NOT, violate energy loss = work done (or more accurately the laws of thermodynamics) when the entire internals of the body is considered.

    BTW this also applies to tossing a ball in the air. This example would not conserve energy. However, compressing a spring and using it to launch a ball which would then come down and land on the spring would (assuming an ideal spring and a vacuum).
  13. Jun 30, 2009 #12
    But work is a scalar quantity.

    You're saying that if you move a weight forward and then back to it's original position, no work will be done because work, being the magnitude of the displacement times the force would be zero,


    surely work was done on the way there
    surely it was done on the way back.

    There is no such thing as negative work, so... what gives?
  14. Jun 30, 2009 #13
    There is most certainly such thing a negative word. A force F is applied in the negative direction (i.e. vector(F)=-F) this force moves something x meters to the right. The work done is -Fx a negative number. And yes, work is a scalar because IT IS ENERGY. That's why its units are joules
  15. Jun 30, 2009 #14
    Saying that no work is done when completing a certain process is exactly identical to saying energy was conserved through out the process.
  16. Jun 30, 2009 #15
    Negative energy?

    Does the negative just indicate what system the energy is "moving" to?
  17. Jun 30, 2009 #16
    work is the amount the energy of the system CHANGED. So there's nothing wrong with negative work.

    P.S. There are many contexts where one will obtain a negative energy without it signifying 'negative energy'. Usually it just means you're in a bound state
  18. Jun 30, 2009 #17
    I feel like I'm probably doing a bad job of explaining things given the amount of confusion. Just see any introductory classical mechanics or thermodynamics book. They'll probably be clearer.
  19. Jun 30, 2009 #18
    but in the eraser on paper example, the energy has changed. A lot.

    I'm tired, and the paper is hot.

    Of course, if you include the paper and myself in the same system, the energy has not changed, which it sounds like you're talking about.

    If what I did to the paper wasn't "work", then what was it?

    What is the name of the process by which I heated up the paper?
  20. Jun 30, 2009 #19
    friction IS NOT A CONSERVATIVE FORCE. Which is what I keep saying. True, friction is ultimately due to microscopic electrodynamic interactions (and electrostatic forces ARE conservative forces) but if you were to consider the system on that microscopic level you really haven't brought the system back to its initial position, the molecules that you heated up now have higher kinetic energys and are who knows where and the system is in an entirely different state even though, to poor human eyes, it looks like we return the system to its original state (the eraser is on the same spot). A conservative force is one, for which, curl(F) = 0 this means:

    [itex]F(x,y,z)\rightarrow \frac{dF_z}{dy}-\frac{dF_y}{dz}=\frac{dF_x}{dz}-\frac{dF_z}{dx}=\frac{dF_y}{dx}-\frac{dF_x}{dy}=0[/itex]

    Gravity, of the form [itex]\frac{GMm}{r^2}[/itex] obeys this, Coulomb's law of the form [itex]\frac{kq_1q_2}{r^2}[/itex] obeys this. However, when we abstract away friction and simply say that it is of the form [itex]\mu N[/itex] friction does not obey this. Therefore, the work done in a closed loop (i.e. your eraser analogy) is not zero. The problem is that by saying friction is of the form [itex]\mu N[/itex] we're really discarding a portion of the system. If we truly considered every molecule making up the pencil, every molecule making up the paper and every molecule of air and treated it that way we would see that energy is conserved. However, by making our friction approximation energy is being lost to 'heat' (although heat is really just kinetic energy of the molecules of the pencil and paper and air)
  21. Jul 1, 2009 #20
    Think of it in terms of action-reaction pairs when it comes to forces. Since dW = F.dx, if I do positive work pushing you through dx with a force F, then you also push on me with -F through the same dx. I have done positive work and expended energy. You have done negative work and gained energy (potential or kinetic, as the case may be).
  22. Jul 1, 2009 #21

    Thanks for all the replies. I guess I am not yet thinking "like a physicist." If I had one car drive around the world (it is a magic car that can drive on water) and return to its original spot, according to physics it did no work. But a car driven linearly for 1 meter did work -- indeed more work than the car that was driven around the world. Even though I can recite the definitions and do the problems, it just seems counter-intuitive to me.

    I guess I have a lot of work (pardon the pun) ahead of me!
  23. Jul 1, 2009 #22
    Don't give up on it Starproj; it will click eventually. maverick_starstrider makes a good points when he suggests getting a good mechanics book. As helpful as a forum can be, it's nice to have a (hopefully) well thought-out and put-together explanation that you can take your time with and know its right.

    Though I have to point out: with the car driven around the world, you will do positive work, even if you end up in the same spot because the frictional force that you work against is not conservative.

    The big distinction here is between conservative and nonconservative forces.

    A conservative force is a special kind of force where the work done while moving along a path depends only on the initial and final points. This, in theory, makes calculations a lot easier (that's why we like working with them so much!). This also means that no work will be done if I start at end in the same place with a conservative force. As an example, take gravity (a prime example of a conservative force). If walk up a flight of stairs and then back down, I have done no net work against the gravitational force. On the trip up I do positive work (and gain gravitational potential energy), but then on the way back down I do negative work (i.e. gravity does the work against me) and lose the potential energy I gained. It all comes out in the wash and net work is zero. Even if I walked up the stairs, took trip down the hall to say "hi" to my friend on the other end, and came back to walk back down, my total work is still zero (on my trip to greet my friend I do no work since my motion is perpendicular to the gravitational force). In one direction I move against the force, in the other I move with it.

    Friction, on the other hand, is nonconservative. On your trip around the world, friction is always acting in opposition to your motion. This means that for every little interval you move, you must do work against friction. And the sum of all of those little positive amounts of work that you have to do on the round-trip add up to a big amount of work that you have to do, and all you get is right back where you started!

    That's the cool thing about conservative forces: they're set up just right (the mathematical stipulation being that they have zero curl) so that if you end where you started, then the little increments of work that you have done in the round trip will be positive and negative in just the right amount to cancel each other out not matter what path you took.

    If you know some vector calculus, you can prove all of this, which definitely helps with understanding it, but if not, just remember that a conservative force is a special kind of force where the work on any round-trip will be zero. And friction is decidedly NOT a conservative force (as maverick_starstrider so pointedly showed in examining the microscopic nature of this force).

    It'll come. Just don't 'force' it ;)
  24. Jul 1, 2009 #23
    By way of an aside, if the race car were to go in water it'd still lose energy since there's water drag and air drag. Think of it this way, a SATELLITE in space (if in a proper orbit) can revolve around the earth and return to its original spot without doing a lick of work (i.e. not expending any energy). This is because gravity is a conservative force. Satellite's do not have to constantly expend energy to return to their starting position. If one could construct an ideal pendulum with zero friction or air resistance it would keep going forever. It would not lose any energy to heat and would constantly return to its starting position with exactly as much energy as it started. Now, of course there's no such thing as a perfect pendulum but hopefully you're starting to get the point.
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