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This question already has an answer here:

Consider the following situation. A block slides on a rough surface. We have already given it an initial velocity. I consider my system to include only the block. Due to friction, the table performs work on the block, thereby reducing its kinetic energy. On the other hand, the block performs no work on the table, since the table surface does not get displaced at all. So, isn't it incorrect to say that work done on the system is the negative of the work done by the system?

If the work done by a system is not always the negative of the work done by the system, then how can both versions of the first law of thermodynamics hold true? The first law of thermodynamics in the physics point of view is: $\Delta U=Q-W,$ and in the chemistry perspective is: $\Delta U=Q+W.$

Both of these seem to be correct, as in the first case we are considering the work done by the system on the surroundings, and in the second case we are considering the work done on the system by external forces. But, if the work done by a system is not equal to the negative of the work done on it, then I have a feeling that the chemistry definition might be a better one. Please clarify.

Kindly note that I am already aware of the different conventions used in physics and chemistry, and I completely understand their intended meanings. My question is not really related to thermodynamics, but something more fundamental. I have just taken the first law of thermodynamics as an example.

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marked as duplicate by ACuriousMind, Gert, Kyle Kanos, Martin, Sebastian Riese Mar 7 '16 at 16:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Does the frictional force on the table due to the block move? $\endgroup$ – Farcher Mar 5 '16 at 7:24
  • $\begingroup$ What do you mean? $\endgroup$ – Newton Mar 5 '16 at 7:27
  • $\begingroup$ if the force moves in the direction of the force then work is done. So if the force on the table due to the block moves then work is done by the force and it does not matter whether or not the table is moving. Just think about you rubbing your hand across a table. $\endgroup$ – Farcher Mar 5 '16 at 7:34
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/72492/2451 , physics.stackexchange.com/q/37904/2451 and links therein. $\endgroup$ – Qmechanic Mar 5 '16 at 8:18
  • $\begingroup$ I am not concerned with the conventions used as I completely understand them. I have a more fundamental question, one not really related to thermodynamics. I have just included the first law as an example. $\endgroup$ – Newton Mar 5 '16 at 8:58
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So, isn't it incorrect to say that work done on the system is the negative of the work done by the system?

It depends on the velocities of the two bodies at the contact point(contact plane). If the two velocities are the same, the two works have the same value and opposite sign. If the velocities are not the same, like your example with friction, there is no such relation. See details in this answer:

How to use the first law of thermodynamics for simple mechanical systems?

If the work done by a system is not always the negative of the work done by the system, then how can both versions of the first law of thermodynamics hold true?

That's a very good question, and the answer is - generally they do not.

What always holds true is this statement:

$$ \text{change of energy of a body} = \text{energy that came as work of other bodies} + \text{energy that came by other means - head conduction, radiation, etc.} $$

To apply this idea to our system, let us introduce notation (in the reference frame of the table):

$E_t$ - energy of table

$\Delta E_t$ - change in energy of the table

$W_{-t}$ - work given to the table (force (due to the block) $\times$ displacement of the table)

$W_{t}$ - work done by the table (force (due to the table) $\times$ displacement of the block)

$Q_{-t}$ - heat given to the table.

The above statement of energy transfer can be then written as

$$ \Delta E_t = W_{-t} + Q_{-t} $$

But this is not the same statement as (!)

$$ \Delta E_t = -W_{t} + Q_{-t}. $$

The last equation is only valid in specific situations in thermodynamics, where $W_{-t}$ is equal to $-W_t$. If friction is present, this is not generally true.

To sum up, in general, the 1st law needs to be formulated as

$$ \Delta U = W + Q $$

where $W$ is accepted work and $Q$ is accepted heat. In general work accepted by body is not equal to work done by body.

In your example, table does negative work on the block, but block does zero work on the table, because the table does not move.

Let's have a look on some concrete numbers. Let's assume the initial kinetic energy of the block is 3 J and 2.7 J ends up in the table, 0.3 J remains in the block.

The work and heat accepted by the table are $W_{-t} = 0$, $Q_{-t}=2.7$

so

$$ \Delta E_t = 2.7~~. $$

The work and heat accepted by the block are $W_{-b} = -3$, $Q_{-b}=0.3$

so

$$ \Delta E_b = -2.7~~. $$

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To answer this question properly, two factors need to be considered.

  1. whether velocity is continuous across the boundary where work is being done
  2. The frame of reference of the observer (since work is a frame-dependent quantity)

If velocity is continuous across the boundary, then the work done by A on B is minus the work done by B on A, irrespective of the frame of reference of the observer. However, if velocity is not continuous across the boundary, the work done by A on B does not have to equal the work done by B on A.

In the case of a gas in a cylinder with a piston, velocity is continuous at the piston face, s0 the frame of reference of the observer doesn't matter. If the observer is stationary relative to the cylinder and the gas is expanding, the work done by the gas on the piston is +W, and the work done by the piston on the gas is -W. If the observer is moving along with the piston, then the piston appears stationary, and the work done by the gas and by the piston are both zero. So, in this moving observer case, what happens to the work? Well, from this frame of reference, the back face of the cylinder is moving backwards, and the gas is doing work on this face (equal to W). The back face of the cylinder is doing equal and opposite work on the gas. So no work is lost.

In the case of a block and a table, velocity is not continuous at the boundary where work is being done, so here, the frame of reference of the observer does matter. From the standpoint of an observer stationary with respect to the table, the table is doing work on the block, but the block is not doing work on the table. There is also a force pushing the block (say at constant speed), so, from the standpoint of the first law, there is no net work being done on the block (W = 0). Of course, considering only the force from the table, the table is doing work on the block (but so is the external force, in the opposite direction).

As reckoned from the frame of reference of an observer traveling with the block, the block is stationary, and the table is moving backwards; the block does work on the table, but the table does no work on the block. Also, as reckoned from this frame of reference, the work done by the external force on the block is likewise zero. So, here again, from the standpoint of the first law, the net work done on the block is zero.

As reckoned by an observer traveling with half the velocity of the block, the block is moving forward and the table is moving backwards at the same speed. So, for this frame of reference the work done by the table on the block is equal and opposite to the work done by the block on the table. In magnitude it is equal to half the work reckoned from the stationary frame. The work done by the external force is also equal to half the value for the stationary frame. So, here again, from the standpoint of the first law, the table does no net work on the block.

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  • $\begingroup$ You are basically implying that from the ground frame, I would get two different first law equations, quite contradictory to one another. $\endgroup$ – Newton Mar 5 '16 at 14:59
  • $\begingroup$ No. I'm going to expand my answer to better clarify what I am saying. Please bear with me. $\endgroup$ – Chet Miller Mar 5 '16 at 15:35
  • $\begingroup$ @kalyan Does my expanded answer work any better for you? If you have any questions, I will be glad to address them. $\endgroup$ – Chet Miller Mar 5 '16 at 19:09

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