Constant velocity problem, where does increase in energy come from?

[Wombat_of_Doom]

Constant velocity problem, where does increase in energy come from?

Let's say that an object is being pushed with a constant force across a surface, and it is moving with a constant velocity. We can use Newton's second law to show that the force of friction is equal to the force of the push.

Considering the object only as the system:
The work done by the push is W = Fpush*x
The work done by the friction is W = -Ff*x
If the force of the push is equal to the force of friction, the net work on the object would be 0.

The object has no change in kinetic energy, no change in potential energy, but should have an increase in thermal energy from the friction that occurs.

If there is no work, how did the total energy of the system increase? What am I missing here?

 

[ZapperZ]

Let's say that an object is being pushed with a constant force across a surface, and it is moving with a constant velocity. We can use Newton's second law to show that the force of friction is equal to the force of the push.

Considering the object only as the system:
The work done by the push is W = Fpush*x
The work done by the friction is W = -Ff*x
If the force of the push is equal to the force of friction, the net work on the object would be 0.

The object has no change in kinetic energy, no change in potential energy, but should have an increase in thermal energy from the friction that occurs.

If there is no work, how did the total energy of the system increase? What am I missing here?

What makes you think that the total energy of the "system" increases? If the "system" is just the mass, there is no increase in energy since v is a constant (no net work done on the mass).

Zz.

 

[jedishrfu]

Let's say that an object is being pushed with a constant force across a surface, and it is moving with a constant velocity. We can use Newton's second law to show that the force of friction is equal to the force of the push.

Considering the object only as the system:
The work done by the push is W = Fpush*x
The work done by the friction is W = -Ff*x
If the force of the push is equal to the force of friction, the net work on the object would be 0.

The object has no change in kinetic energy, no change in potential energy, but should have an increase in thermal energy from the friction that occurs.

If there is no work, how did the total energy of the system increase? What am I missing here?

This is pretty much the same as lifting an object up from the ground such that it moves at constant velocity. Your lifting force is applied over a distance giving the object potential energy. Where did the energy come from? It came from chemical reactions giving energy to your muscles to in turn lift the object.

 

[Wombat_of_Doom]

If the object has increased in temperature due to friction, hasn't it gained thermal energy?

If Kinetic energy stays constant, potential energy stays constant, and thermal energy increases, isn't that a total gain in energy for the system?

 

[Wombat_of_Doom]

This is pretty much the same as lifting an object up from the ground such that it moves at constant velocity. Your lifting force is applied over a distance giving the object potential energy. Where did the energy come from? It came from chemical reactions giving energy to your muscles to in turn lift the object.

I've already accounted for that chemical energy when I considered the work done by the push.

 

[sophiecentaur]

If the object has increased in temperature due to friction, hasn't it gained thermal energy?

If Kinetic energy stays constant, potential energy stays constant, and thermal energy increases, isn't that a total gain in energy for the system?

That 'gained energy' has to come from the source that's doing the work. You always need to look for something you missed in these questions. In this case, the system you described is incomplete.

 

[Wombat_of_Doom]

That 'gained energy' has to come from the source that's doing the work. You always need to look for something you missed in these questions. In this case, the system you described is incomplete.

Right. The gained energy due to the work of the push is negated by the lost energy due to the work of the ground via friction. Both forces are equal, both displacements are equal. Yet there is still a gain in thermal energy. Unless you are suggesting that the work done by the push is greater than the work done by the force of friction.

I understand that I'm missing something, but nobody has really told me what. Either explain why these two sources of work are not equal, find me a third source of work, or explain to me why there isn't actually a gain in energy of the object.

 

[sophiecentaur]

Two sources? You put energy into the system in the form of work (one source) and that work turns up as thermal energy - PE and KE being unchanged. Isn't that enough? What else are you looking for?

 

[jedishrfu]

Right. The gained energy due to the work of the push is negated by the lost energy due to the work of the ground via friction. Both forces are equal, both displacements are equal. Yet there is still a gain in thermal energy. Unless you are suggesting that the work done by the push is greater than the work done by the force of friction.

I understand that I'm missing something, but nobody has really told me what. Either explain why these two sources of work are not equal, find me a third source of work, or explain to me why there isn't actually a gain in energy of the object.

By having a frictional force you are by definition having the object rub against something and that produces heat. You could look at it on the molecular level and perhaps that would help you understand with molecules pushing against molecules and hotter ones moving faster...

 

[Wombat_of_Doom]

Isn't there work done on the object by the push as well as by the friction? If you want to consider all the forces as one net force, then the net force would be zero, making the net work zero.

 

[sophiecentaur]

Isn't there work done on the object by the push as well as by the friction? If you want to consider all the forces as one net force, then the net force would be zero, making the net work zero.

I think you're trying to count the same work twice here. The work done against friction is what counts in the system you're looking at. There is only one force and one distance involved. Would you talk about a car doing work against the engine whilst it's being accelerated? I don't think so. So where is the difference just because it's friction that's involved? If you are consistent then I think there is no problem.

 

[sophiecentaur]

I just thought some more about this and it may be that your worry is to do with the energy lost within the source. If you take an electrical analogue, source impedance will always involve some extra energy dissipation - in addition to that lost in the circuit itself.

 

[Wombat_of_Doom]

I think you're trying to count the same work twice here. The work done against friction is what counts in the system you're looking at. There is only one force and one distance involved. Would you talk about a car doing work against the engine whilst it's being accelerated? I don't think so. So where is the difference just because it's friction that's involved? If you are consistent then I think there is no problem.

This is how I'm seeing it, so help me out with my problem. The force diagram for my object would have two force vectors (ignore gravity and normal because they are perpendicular to displacement). The force of the push, acting to the right, and the force of friction, acting to the left. The push acts in the direction of the displacement, so positive work. The friction acts opposite the displacement, so negative work. The ground is not a part of my system, so it is not an internal force like your car/engine example.

 

[sophiecentaur]

Friction is not a source of energy, whatever you say about "negative work". The loss involved with the friction is just one step in the transfer of your input work to thermal.
Perhaps your problem is that you are equating a force diagram with an energy diagram. 'Thermal' can't be shown on a force diagram so you can't use a force diagram except to show that work in is the work against (or by) the friction. Thereafter, the work done / energy lost with the friction appears as thermal energy.

I don't understand your comment about the "ground", I'm afraid.

 

[256bits]

This is how I'm seeing it, so help me out with my problem. The force diagram for my object would have two force vectors (ignore gravity and normal because they are perpendicular to displacement). The force of the push, acting to the right, and the force of friction, acting to the left. The push acts in the direction of the displacement, so positive work. The friction acts opposite the displacement, so negative work. The ground is not a part of my system, so it is not an internal force like your car/engine example.

If I may add,
If you were pushing a block to the right on a table, with a force F a distance x, then your work would be Fx. Normally they teach that friction also is a work force and that work is Ffx, just to make it easier to understand and acquire original concepts.
But at the intersection between the block and table there are 2 friction forces, one of the block pushing on the table to the right and another of the table pushing to the left. Since the table is not moving, is Ff table really causing any work.

You can see relate that to the table moving underneath the block and you holding the block in one position. Is the force F that you push on the block doing any work?

But that is all a macroscopic viepoint.

Actually, if one goes down to the molecular level it is the bonds between surface atoms of the block and table that stretch and release which would then cause the atom or molecules to vibrate and acquire thermal energy. That is a simplistic explanation. People still do not fully understand exactly what is friction.

Getting back to your question, even though you have a friction force opposing the block's motion, and you say F x - Ff xf = 0 but still the block acquires thermal energy. In which case, to balance energy, Ff xf <F xf , so if Ff < F the block would accelerate and not have a constant velocity. Or xf < x which is the only alternative. So the premise That F x = Ff xf is false. In other words xf < x, since we do know that F = Ff.

What you need to state is that F = Q, if I may use Q here, but just to signify ONLY that there is somehow a heat gain. Your force of pushing the block on the table results in a thermal energy increase of the block.

As an analogy, a piston pushing down on a gas in a cylinder has a force opposing motion from the gas, but as the gas is compressed its temperature increases. Here we do not say that F - Fgas = 0 so work is zero, but the gas still acquires thermal energy. We say that the work input into the gas results in an increase in the internal energy of the gas.

Hope that makes sense.

 

[Jupiter6]

Let's say that an object is being pushed with a constant force across a surface, and it is moving with a constant velocity. We can use Newton's second law to show that the force of friction is equal to the force of the push.

Considering the object only as the system:
The work done by the push is W = Fpush*x
The work done by the friction is W = -Ff*x
If the force of the push is equal to the force of friction, the net work on the object would be 0.

The object has no change in kinetic energy, no change in potential energy, but should have an increase in thermal energy from the friction that occurs.

If there is no work, how did the total energy of the system increase? What am I missing here?

I'm certainly no physics major but I'm wondering:

How does Newton's 2nd apply to this situation?

How is this object moving across a surface if the push force is equal to the friction force? The push force would have to be greater than the opposing friction force in order for it to move.

 

[sophiecentaur]

I'm certainly no physics major but I'm wondering:

How does Newton's 2nd apply to this situation?

How is this object moving across a surface if the push force is equal to the friction force? The push force would have to be greater than the opposing friction force in order for it to move.

Nope if the two forces are equal and opposite, the velocity will remain constant. That's Newton One. You are thinking subjectively and not applying basic principles. The 'push' force is already taking care of the friction force and you don't need 'extra' any more than if the object were out, drifting in deep space. Since Newton, we all know that things don't "naturally slow down".

 

[Jupiter6]

Nope if the two forces are equal and opposite, the velocity will remain constant. That's Newton One. You are thinking subjectively and not applying basic principles. The 'push' force is already taking care of the friction force and you don't need 'extra' any more than if the object were out, drifting in deep space. Since Newton, we all know that things don't "naturally slow down".

Considering the object and surface as our frame of reference, neither is moving with respect to each other. "Drifting through space" implies another reference frame which is not something I would assume from the OP's question considering we need gravity from some large body to supply a gravitational attraction to produce a coefficient of friction.

Please tell me how I'm failing to apply basic principles.

EDIT: Forget it Sophie. Looking at your other posts, it's apparent you never offer anyone real-world advice, yet you get off on confusing people. Go smoke another joint.

 

[sophiecentaur]

Considering the object and surface as our frame of reference, neither is moving with respect to each other.

? Do you really mean that? I thought the scenario was an object moving across a surface against friction i.e. one relative to the other.

Just apply Newton1 to the problem. If there is no NET force then an object will maintain its momentum. When push and friction are equal, the object will maintain its speed (whatever its value - including zero) for ever.

 

[russ_watters]

I think you're trying to count the same work twice here.

No, s/he is counting the work done by the friction -- producing heat -- but not counting the work done by the force. That's the error.

The net work on the box is zero, but work is done by the force and by friction. The input comes from whatever is applying the force and the output comes from friction: Win-Wout=0

 

[256bits]

No, s/he is counting the work done by the friction -- producing heat -- but not counting the work done by the force. That's the error.

The net work on the box is zero, but work is done by the force and by friction. The input comes from whatever is applying the force and the output comes from friction: Win-Wout=0

Then we are back to the original question of what is causing an increase in temperature of the block.
Win = ΔU an increase of the temperature of the mnating surfaces , as per my first post.
Otherwise conservation of energy is violated.

 

[jedishrfu]

I'm certainly no physics major but I'm wondering:

How does Newton's 2nd apply to this situation?

How is this object moving across a surface if the push force is equal to the friction force? The push force would have to be greater than the opposing friction force in order for it to move.

This is a common issue with students learning physics. When there's no force on an object, it either remains still or moves in a straight line at constant speed. This is also true when there are forces that cancel out. In this case, the friction force is dependent on the speed of the object. If you push harder then the object moves faster and should accelerate ie go faster and faster but because the frictional force pushes back at the same strength thus we get a constant speed for the object.

So when friction is present a continuous push results in an object moving at constant speed and when the push changes strength or stops then the object changes speed or stops.

Another feature of frictional forces is that there's a static frictional force to overcome first and then a lower moving frictional force pushing back. All of this gets too complicated for beginning students of physics so the problems skirt the issue and hope you won't notice but will understand and accept from your commonsense experience.