# Displacement and Work

1. Jun 29, 2009

### Starproj

Hi,

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. Jun 29, 2009

### Archosaur

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.

Make sense now?

3. Jun 29, 2009

### maverick_starstrider

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...

4. Jun 29, 2009

### Archosaur

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.

5. Jun 29, 2009

### maverick_starstrider

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.

6. Jun 29, 2009

### diazona

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
$$W = \int \vec{F}\cdot\mathrm{d}\vec{s}$$
For conservative forces, it just happens that the result of adding up all those little bits of work depends only on the net displacement.

7. Jun 30, 2009

### infomax

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

8. Jun 30, 2009

### cipher42

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.

9. Jun 30, 2009

### Topher925

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.

10. Jun 30, 2009

### Archosaur

So, work is energy spent against a field force, such that energy is stored?

aka, lifting a weight, or pulling back a rubber band?

11. Jun 30, 2009

### maverick_starstrider

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).

12. Jun 30, 2009

### Archosaur

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,

but

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

There is no such thing as negative work, so... what gives?

13. Jun 30, 2009

### maverick_starstrider

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

14. Jun 30, 2009

### maverick_starstrider

Saying that no work is done when completing a certain process is exactly identical to saying energy was conserved through out the process.

15. Jun 30, 2009

### Archosaur

Negative energy?

Does the negative just indicate what system the energy is "moving" to?

16. Jun 30, 2009

### maverick_starstrider

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

17. Jun 30, 2009

### maverick_starstrider

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.

18. Jun 30, 2009

### Archosaur

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?

19. Jun 30, 2009

### maverick_starstrider

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:

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

Gravity, of the form $\frac{GMm}{r^2}$ obeys this, Coulomb's law of the form $\frac{kq_1q_2}{r^2}$ obeys this. However, when we abstract away friction and simply say that it is of the form $\mu N$ 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 $\mu N$ 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)

20. Jul 1, 2009

### cipher42

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).