the definition of 'work'

quantum quack · 06-02-2005

in a river I place a rock that deflects the fast flowing water, is the rock doing work?

if so how is it's energy use conserved?

the question is really about how a seemingly static object can perform work or at least a function like deflecting a torrent of water with out actually expending any recordable energy. Yet it provides a force that deflects the energy of the water [in this example]

where does the object get the energy from to deflect water?

Boerseun · 06-02-2005

Quote:

Originally Posted by quantum quack

in a river I place a rock that deflects the fast flowing water, is the rock doing work?

if so how is it's energy use conserved?

the question is really about how a seemingly static object can perform work or at least a function like deflecting a torrent of water with out actually expending any recordable energy. Yet it provides a force that deflects the energy of the water [in this example]

where does the object get the energy from to deflect water?

Yes, the rock is doing work.
The result of this work is the ripples of water you'd see trailing the rock, much like a boat's wake.
The rock itself is not the agent of the work being done, but is just there, by accident, so to speak. The actual effort is in pressure from the water trying to get the rock to overcome the friction between itself and the riverbed.
The friction between the rock and the the riverbed depends on the weight of the rock, and it's texture. If the rock was very light, the stream won't have much of an effort to wash it away. But being a normal rock, I assume it would be considerably heavier than water, which means you need a fairly fast flowing stream to push it. Unless both the riverbed and the rock are very, very smooth - then once again the friction will be reduced.
Now - say the rock is so heavy and coarse so as not to move at all? Is there any work being done?
Yes.
But not readily visible, and actually quite abstract. The force of the water pushing against the rock (of which the ripples is the evidence) is transmitted to the riverbed, which is fixed to the Earth's crust. So, in effect, the small stream pushing against the small rock is having an effect on the Earth itself. You might say that in ten kazillion years, when the Earth comes to an end, the (floating) continental plate experiencing the effect of your rock, will be two inches away from where it would have been, had you not thrown a rock in the stream.
Case in point - it has been calculated that the planet Jupiter will in 4 billion years be two feet short of where it was supposed to be, had Voyager 2 not stolen some of its angular momentum on its flyby in the eighties. Amazing.

TeleMad · 06-04-2005

Quote:

quantum quack: in a river I place a rock that deflects the fast flowing water, is the rock doing work?

Let's see. Since the flow of water is being diverted, then the water molecules are having their direction of motion changed ... by the rock. A change in direction is an acceleration, so the rock is causing the water molecules to accelerate. And it takes a force to cause acceleration, so the rock must be exerting a force against the water molecules.

This is a reaction force, according to Newton's third law (for every action force there is an equal and opposite reaction force). The water is in motion and is striking the rock, exerting an action force against it (but doing no observable work, since the rock doesn't move to any measurable degree). The rock is in turn exerting a reaction force against the water molecules.

Since the molecules of water are in a fluid state, they can be considered to be nearly independent: at least more independent than the molecules that make up the solid rock, which all act as a single unit. So it is the water molecules that experience the (greater) acceleration.

So far, we have the water molecules exerting a force against the rock but doing no work (since the rock doesn't move), and the rock exerting a reaction force against the water molecules and causing them to accelerate.

That all seems straightforward. But the original question still remains unanswered: does what the rock is doing count as work?

As alluded to above, work requires not just a force being exerted against an object, but also for the object having the force exerted against it to be moved a distance as a result of the force. Now, the rock is passively deflecting the water molecules off into a different direction - it is accelerating them by means of a reaction force - but does that actually count as moving the water molecules through a distance because of the force exerted on them? I am not sure.

quantum quack · 06-04-2005

Do you think the rock expends energy to do what it does?
If so where does the energy come from and how is it conserved?

C1ay · 06-04-2005

IMO the rock does no work, the water does by eroding the rock over time. With it's force it removes particles from the rock and moves them a distance as a result. The rock only reacts to the energy of the water, not by any energy of it's own. It is like a tennis ball dropped on the concrete. The rebound of the ball is a result of the balls energy, not the concrete's energy.

TeleMad · 06-05-2005

Because I was unsure, I asked at another board. Apparently no actual physicists responded, and there were replies that supported "yes" and some that supported "no".

Here's an idea I overlooked that might help. Consider the relative motion between the water and the rock from the point of view of stationary water and moving rock (Galileian relativity). Then we have the rock plowing through water molecules and pushing them out of the way. This would be in line with the rock having performed work on the water.

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I can imagine where the energy to do work - if such is being done - could be coming from. We would have to again look at the atomic scale where the force involved is the electrostatic force of electron shells in the water molecules and in the rock repelling each other when they come close enough to one another. Consider this again from the point of view of moving water and a stationary rock. The electrons of an approaching water molecule partially displace the electrons of the atoms in the rock's surface (that electron density can be redistributed by electrical forces is well known: an example is what occurs when you stick a balloon that has been rubbed against your hair to a wall. The negative charges on the balloon pushes the electrons in the atoms in the wall's surface towards the opposite side). This distortion of the electron distribution generates a sort of minute elastic potential energy, analogous to compressing a spring. As with a spring, the system restores itself to its original state, thereby returning the energy that was stored in it.

Erasmus00 · 06-05-2005

If you place a rock in a fast moving stream, the question of wether the rock does work is easiest analyzed by thinking in terms of energy. Originally, the water molecules are fast moving, so the kinetic energy of a water molecule is fairly high. After colliding with the rock, the water molecules are now moving much more slowly, so their energy has in fact been decreased. Where does this energy go? The rock heats up a bit(in other word, the rock molecules start vibrating), and that heat gets carried into the river bed its anchored in.
In general, when you have friction processes like this (rock in a stream, ball sliding to a stop) the energy that looks "lost" becomes heat.
Now, from a galilean relativistic point of you, this situation should by dynamically equivalent to a rock moving through a still stream. Now, the rock must do work, becasue the water molecules start speeding up, gaining energy. Where does that energy come from? Well, either there are no outside forces, and in fact the rock starts to slow down, or else the rock has another power source (you pushing it, for instance). So, which system is gaining energy and which is losing energy (hence which does work, rock or water) totally depends on your frame of reference. Just another of those tricky things about the whole energy/work concept.
-Will

quantum quack · 06-05-2005

Facinating responses guys thanks.
If we take a steel gallow frame and hang a 10 kg weight from a cable is the frame doing work?
I am assuming the term work means "to expend energy".
It's sort of like looking at a steel bar standing on it's end and asking is it expending energy to sustain it's rigidity?
I tend to think that the answer is that the energy expended is also conserved with in the bar system. [at an atomic level] that the forces within the bar generate the energy and absorb the energy in doing that generation. a cycling of energy so to speak.

it is an interesting question though, as to how to correctly assess the energy situation of our rock in a river, I must admit.

infamous · 06-05-2005

Quote:

Originally Posted by C1ay

IMO the rock does no work, the water does by eroding the rock over time. With it's force it removes particles from the rock and moves them a distance as a result. The rock only reacts to the energy of the water, not by any energy of it's own. It is like a tennis ball dropped on the concrete. The rebound of the ball is a result of the balls energy, not the concrete's energy.

I agree C1ay; The rock is doing no work at all, even the ripples that are created are the effect of energy in the water flow, therefore the water is doing all the work.