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Displacement and Work 
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#1
Jun2909, 02:04 PM

P: 18

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
Jun2909, 03:25 PM

P: 328

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? 


#3
Jun2909, 03:54 PM

P: 1,164

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
Jun2909, 09:30 PM

P: 328

Displacement and Work
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
Jun2909, 10:29 PM

P: 1,164

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
Jun2909, 11:32 PM

HW Helper
P: 2,155

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 nonmathematical 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. 


#7
Jun3009, 08:26 AM

P: 8

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
Jun3009, 08:55 AM

P: 49

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
Jun3009, 09:31 AM

P: 1,672

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
Jun3009, 11:42 AM

P: 328

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
Jun3009, 02:57 PM

P: 1,164

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
Jun3009, 03:20 PM

P: 328

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
Jun3009, 07:11 PM

P: 1,164

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
Jun3009, 07:12 PM

P: 1,164

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



#15
Jun3009, 07:38 PM

P: 328

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


#16
Jun3009, 07:41 PM

P: 1,164

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
Jun3009, 07:44 PM

P: 1,164

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
Jun3009, 07:45 PM

P: 328

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? 


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