Question about the physics term "work"

[Physics is awesome]

I have a question about work. If Newtons 3rd law of motion states there is an equal and opposite reaction for any force. Now if for example I apply 500 Newtons of force on my dresser and move it 5 meters I used 2500 joules of energy. Now because the dresser is equally putting that much force in return for the same distance does this also mean the dresser used 2500 joules of energy resisting my force? So I used 5000 joules of energy?

 

[QuantumQuest]

I have a question about work. If Newtons 3rd law of motion states there is an equal and opposite reaction for any force. Now if for example I apply 500 Newtons of force on my dresser and move it 5 meters I used 2500 joules of energy. Now because the dresser is equally putting that much force in return for the same distance does this also mean the dresser used 2500 joules of energy resisting my force? So I used 5000 joules of energy?

Initially the dresser is at rest. Now, when you exert a force on it, it won't get into motion till the force you exert overcomes the resultant reactive force. From this point on the dresser is in motion - i.e. the dresser does not exert an equal force anymore, in the direction of the force you exert and the work you consume becomes kinetic energy minus the work that is consumed by the resultant reactive force (i.e. friction). It is always a good idea to draw a two-axes diagram in order to see what are the forces in both axes and see it in more detail.

 

[russ_watters]

Now if for example I apply 500 Newtons of force on my dresser and move it 5 meters I used 2500 joules of energy. Now because the dresser is equally putting that much force in return for the same distance does this also mean the dresser used 2500 joules of energy resisting my force?

Kind of. It applied that force to you in the opposite direction you did. But it also applied a force in the same direction, against the ground (friction).

So I used 5000 joules of energy?

Huh? You just said you used 2500N (correctly)!

 

[Chestermiller]

You did 2500 J work on the dresser, and the dresser did -2500 J of work on you. So??

 

[sophiecentaur]

You did 2500 J work on the dresser, and the dresser did -2500 J of work on you. So??

It’s better to include the Momentum change. Pushing something with your feet firmly on the floor involves a near zero movement of your feet / Earth. So Force X Distance is Zero.
Momentum is the only conservation law that you can lean on in this kind of argument.

 

[Chestermiller]

It’s better to include the Momentum change. Pushing something with your feet firmly on the floor involves a near zero movement of your feet / Earth. So Force X Distance is Zero.
Momentum is the only conservation law that you can lean on in this kind of argument.

Are you saying that what I said is incorrect, to wit: the work that the dresser does on you is equal in magnitude and opposite in sign to the work you do on the dresser?

 

[sophiecentaur]

Are you saying that what I said is incorrect, to wit: the work that the dresser does on you is equal in magnitude and opposite in sign to the work you do on the dresser?

If you consider where the Energy ends up then there is no reason to suppose that the Energy from the muscles should be shared equally.
You will say it’s all a matter of reference frames, no doubt. But Newton 3 only refers to force and not work.

 

[Chestermiller]

If you consider where the Energy ends up then there is no reason to suppose that the Energy from the muscles should be shared equally.
You will say it’s all a matter of reference frames, no doubt. But Newton 3 only refers to force and not work.

I don't follow. Isn't the work I do on the the dresser equal to force I apply times the distance over which I apply it? And isn't the work that the dresser does on me equal to the force that it applies on me times the distance over which the force is applied (with a minus sign because the force and displacements are in opposite directions)? We seem to be talking about two different things.

 

[sophiecentaur]

We seem to be talking about two different things.

Yes, I think we are. The strict definition of Work, which is Force times Distance would suggest that you are right but the share of the energy is not equal. Perhaps it's just a bad idea to try to resolve this question. The change in energy of two objects, mutually repelling is not necessarily equal so saying that the Work Done is equal is not really relevant. It raises a question without answering it satisfactorily. The very fact that we (and so many people in history) have this conversation from time to time is because there is a perceived paradox and it makes people uneasy.
The earlier posts in which the Energy is discussed show the problem of going down this particular road because it leads to implications about Energy Conservation that are not true.

 

[Mister T]

The problem occurs when considerations of work done are generalized to considerations of energy. There is nothing wrong, for example, with saying that 2500 joules of work is done by the boy on the dresser, which is of course equivalent to saying that -2500 joules of work is done by the dresser on the boy. This is nothing more than saying that when 2500 joules of work is done on the dresser, -2500 joules of work is done by the dresser.

Problems occur when we make statements about energy transfer, like saying that 2500 joules of energy was transferred to the dresser by the boy. Such a statement is valid only if the dresser can be modeled as a particle, unable to absorb internal energy and undergo a temperature change. For example, if the dresser moves at a steady speed the 2500 joules of work done by the boy on the dresser may not constitute a transfer of 2500 joules of energy to the dresser, some of that 2500 joules may be transferred to the dresser, increasing its temperature, but the rest may be transferred to the floor raising its temperature. And statements such as the work done "goes into heat" are also incorrect, as in this case there is no transfer of heat, the process is adiabatic.

 

[jbriggs444]

The change in energy of two objects, mutually repelling is not necessarily equal

The work done by each of two objects on the other when interacting with a contact force (e.g. a normal force) where the contacting surfaces are not in relative motion will always be equal and opposite.

Kinetic friction and non-contact forces present opportunities for the two energy changes to differ.

 

[sophiecentaur]

The work done by each of two objects on the other when interacting with a contact force (e.g. a normal force) where the contacting surfaces are not in relative motion will always be equal and opposite.

Kinetic friction and non-contact forces present opportunities for the two energy changes to differ.

We sort of dealt with that. Whilst it is numerically true, it does not imply anything about the Energy.
Conserving Momentum is sacrosanct and both the contact times and distances are the same. Work Done is not always an important quantity and it isn't particularly relevant in many circumstances. Friction and "non contact Forces" do not affect the general principle.
Work done is fine for ' one sided' processes like a ballista or a gun but I wouldn't know where to start to do an impact calculation by replacing Impulse with Work. I wouldn't even like to do the sums on an artillery shell by choosing the shell's reference frame, despite the fact that Work can be calculated in both directions.

 

[jbriggs444]

Whilst it is numerically true, it does not imply anything about the Energy.

It does not imply anything. It states it flat out. Work is a transfer of energy.

 

[sophiecentaur]

It does not imply anything. It states it flat out. Work is a transfer of energy.

Yes, you are stating something that is more like a mathematical identity which needs some context to have physical meaning. "Transfer of Energy" from what to what? Starting with a coiled spring between the Earth and a stone, where does the energy end up? Your answer must satisfy momentum conservation.
We are still arguing in quadrature. The fact that the question is asked so often, serves to demonstrate that the Work Done idea is not always very helpful and needs to be helped along.

 

[jbriggs444]

"Transfer of Energy" from what to what? Starting with a coiled spring between the Earth and a stone, where does the energy end up?

One can examine the energy transferred by work at each interface.

Pick a reference frame. The frame in which the Earth is initially at rest is convenient.

The interface between spring and Earth is [at least approximately] immobile. No energy is transferred between the two. The interface between spring and stone moves in the direction of the force on the stone. Energy is transferred to the stone. The stone gains energy from the interaction. Energy is transferred from the spring. The spring loses energy from the interaction.

If you prefer to use a different frame, that can be done. It turns out that the net energy loss in the spring is a classical invariant. The energy changes in Earth and stone are not invariant, but will sum to a figure that is invariant and is equal and opposite to the energy change in the spring.

 

[sophiecentaur]

Pick a reference frame. The frame in which the Earth is initially at rest is convenient.

That's the one I would choose.

the net energy loss in the spring is a classical invariant

Absolutely. Always.
The point I am making is that there is confusion about this. If, like you, people follow the Maths scrupulously, then the conclusion is what it is. The bare facts of Work being done both ways is just not helpful for anyone who's mind strays from the 'literal'. Rather than insisting that the definition is what it is, why not help the situation along if it doesn't actually involve introducing an error? What I am introducing is no more than stating the sort of problem that rocket engines encounter at low speeds. All the energy goes into the ejecta and virtually none goes into the rocket at the start.

 

[A.T.]

I have a question about work. If Newtons 3rd law of motion states there is an equal and opposite reaction for any force. Now if for example I apply 500 Newtons of force on my dresser and move it 5 meters I used 2500 joules of energy. Now because the dresser is equally putting that much force in return for the same distance does this also mean the dresser used 2500 joules of energy resisting my force? So I used 5000 joules of energy?

Look at the exact vector definition of work, via the dot product: Since force and displacement are opposite, the work done by the dresser on you is negative.

 

[A.T.]

Your answer must satisfy momentum conservation.

Not really what the OP asks about.

 

[jbriggs444]

The bare facts of Work being done both ways is just not helpful for anyone who's mind strays from the 'literal'. Rather than insisting that the definition is what it is, why not help the situation along if it doesn't actually involve introducing an error?

I think I understand your point better now. At the risk of oversimplifying: If the concept of work is confusing, there are two possible approaches. One is to clarify the confusion. The other is to use a different concept.

 

[Mister T]

The work done by each of two objects on the other when interacting with a contact force (e.g. a normal force) where the contacting surfaces are not in relative motion will always be equal and opposite.

This is a perfectly valid dynamical relation.

Kinetic friction and non-contact forces present opportunities for the two energy changes to differ.

Suppose a person on roller skates launches himself into motion across the floor by pushing on a wall. This would seem to qualify as an example of what you mention above, but in what way are energy changes equal?

 

[jbriggs444]

Suppose a person on roller skates launches himself into motion across the floor by pushing on a wall. This would seem to qualify as an example of what you mention above, but in what way are energy changes equal?

Seen as a contact force: The energy imparted by the wall on the surface of the hand is zero. The energy imparted by the surface of the hand on the wall is zero.

Seen as a force at a distance (mediated by the non-rigid arm): The energy imparted by the wall on the body is non-zero. The energy imparted by the body on the wall is zero.

Edit: It is intuitively jarring to imagine the static wall doing work on the person. That is because the arm has been abstracted away into the "mediated by" clause and is the energy source for the work done on the body.

 

[A.T.]

Suppose a person on roller skates launches himself into motion across the floor by pushing on a wall. This would seem to qualify as an example of what you mention above, but in what way are energy changes equal?

The person does zero work in the static wall. The static wall does zero work on the person. No energy is transferred between the static wall and the person..

 

[PeterO]

Are you saying that what I said is incorrect, to wit: the work that the dresser does on you is equal in magnitude and opposite in sign to the work you do on the dresser?

BUT work and energy are both scalar quantities - should they have a sign?

 

[PeterO]

Look at the exact vector definition of work, via the dot product: Since force and displacement are opposite, the work done by the dresser on you is negative.

But the result of a vector dot product is a scalar. Can it ever be negative? I thought scalars only had magnitude.

 

[A.T.]

But the result of a vector dot product is a scalar. Can it ever be negative?

Look at the definition of the dot product.

 

[PeterO]

Look at the definition of the dot product.

I did look at the definition, and it confirms the answer is a scalar - therefore has no direction (and that's where the positive and negative comes in) Even if the "mathematical" calculation has a negative (or positive) sign on the numerical answer, the sign is ignored for a scalar is it not.

 

[A.T.]

the sign is ignored for a scalar is it not.

Look at the definition of a scalar.

 

[PeterO]

Look at the definition of a scalar.

OK I will re-word that.
Can you give an example of a scalar with anything other than magnitude?

 

[A.T.]

...with anything other than magnitude

Too vague. Do you mean a sign?

 

[Chestermiller]

But the result of a vector dot product is a scalar. Can it ever be negative? I thought scalars only had magnitude.

What if the vectors in the dot product are pointing in opposite directions?

 

[PeterO]

What if the vectors in the dot product are pointing in opposite directions?

The mathematics (arithmetic really) will have a negative sign, but when the answer is interpreted for to the situation (a scalar quantity) that sign is ignored.

Take the case of a mass moving in a circle. On opposite sides of the circle it is moving in opposite directions, but it does not have positive Kinetic energy on one side, and negative on the other. It just has kinetic energy; neither positive nor negative (because kinetic energy, like Work, is a scalar).

 

[A.T.]

The mathematics (arithmetic really) will have a negative sign, but when the answer is interpreted for to the situation (a scalar quantity) that sign is ignored.

No, the sign is interpreted as the direction of energy transfer. When A is doing negative work on B, then energy is transferred from B to A.

 

[Chestermiller]

The mathematics (arithmetic really) will have a negative sign, but when the answer is interpreted for to the situation (a scalar quantity) that sign is ignored.

Take the case of a mass moving in a circle. On opposite sides of the circle it is moving in opposite directions, but it does not have positive Kinetic energy on one side, and negative on the other. It just has kinetic energy; neither positive nor negative (because kinetic energy, like Work, is a scalar).

The sign is not ignored. If A is doing negative work on B (i.e., with the sign included), it just means the B is doing positive work on A.

In thermodynamics, we encounter negative work all the time when we write $W=\int{PdV}$, where W is the work done by the system on the surroundings. If the gas is expanding, we have that the system is doing positive work on the surroundings, which, from the first law of thermodynamics, $\Delta U=Q-W$, means that the positive work that the system does is tending to decrease its internal energy U (and temperature). If the gas is being compressed by the surroundings, we have that the system is doing negative work on the surroundings, which, from the first law of thermodynamics, means that the negative work that the system does on the surroundings (i.e., the surroundings are doing positive work on the system) is tending to increase its internal energy (and temperature).

Just try doing thermo without the concept of work being allowed to be negative, and see how confused you will get.

 

[PeterO]

No, the sign is interpreted as the direction of energy transfer. When A is doing negative work on B, then energy is transferred from B to A.

I agree the negative sign that the arithmetic gives you can be used to interpret some details of the energy changes, but cannot agree that there is negative work. Work is a scalar quantity, and the idea of positive and negative are simply not relevant.

 

[Chestermiller]

I agree the negative sign that the arithmetic gives you can be used to interpret some details of the energy changes, but cannot agree that there is negative work. Work is a scalar quantity, and the idea of positive and negative are simply not relevant.

Then how do you account for our approach in thermodynamics, where it is extremely relevant (and automatically gives you the correct answer)?