First law of thermodynamics, pulling an object over a surface

Question

Consider a box pulled over a surface, where friction is involved. Clearly heat is generated here, and so one might possibly consider using the first law. The first law says that $$\Delta U=Q-W$$

The law requires a system to be determined, and I say it consists of the box and the surface. If we ignore the effects on the external environment, this is a closed system ( or just insulate the whole thing, whatever). The work done by the system is $W = F_{\mu}d$, when friction $F_\mu$ acts over a distance $d$. Is it now correct to say that $$\Delta U=Q-F_{\mu}d$$

or $$0 = Q - F_{\mu}d$$

Answer 1

If the box and the surface are the system, then the air surrounding the box and surface, and whatever lies below the surface, are part of the surroundings. Since the box does not "move itself" the force that pushes the box should be considered as part of the surroundings.

A closed system is one that does not exchange mass with its surroundings. But it can exchange energy in the form of work and/or heat. So in this case, the external agent applying a force of $F_{ext}$ does work of $F_{ext}d$ on the system and the system can transfer heat to the surroundings (i.e., it is not thermally insualted).

In order to apply the first law, not only do you need to define the system, but you also need to define the initial and final equilibrium state of the system. To that end, let's assume the initial equilibrium state is with the box at rest with the system (box and surface) in thermal equilibrium with the surroundings which we will consider to be at constant temperature (i.e., the surroundings is a thermal reservoir). The final equilibrium state after the box has moved a distance $d$ is the box is again at rest, and the box and surface are again thermal equilibrium with the surroundings.

Since the box begins and ends at rest the change in kinetic energy is zero and from the work energy theorem that means the net work done on the system is zero. In this case, the negative work done by friction equals the positive work done by the external force, or $F_{u}d=F_{ext}d$. Since the friction work is between the box and surface, that work is internal to the system.

In effect, the negative internal work done by friction takes the energy transferred to the box by the external force and temporarily stores it as internal energy of the box/surface system. I say temporarily because unless the box and surface are thermally insulated from its surroundings, the increase in system temperature relative to the surroundings will result in heat $Q$ being transferred from the system to the surrounding until thermal equilibrium is reestablished.

Thus, under the above conditions and assumptions, the first law gives us

$$\Delta U=Q-F_{ext}d=0$$

However, since $F_{ext}=F_u$

$$0=Q-F_{u}d$$

So your conclusion would be correct under the conditions described above.

Hope this helps.

Answer 2

Let system be surface plus box. First suppose that the situation you describe involves some external agent producing the force which makes the box move. This external agent does work $F_\mu d$ on the system. Meanwhile there is no heat transfer, so $Q=0$. The system's internal energy consequently goes up by $\Delta U = F_\mu d$.

If instead you had a mind a completely isolated system, so $\Delta U = 0$, then why is the box moving? It could be that it was set in motion at some moment, and now the system is returning to equilibrium. In this case you have $Q = 0$ and $W=0$. No work is being done on the system, but internally to the system (of box plus surface), kinetic energy of motion of the box is being converted into random motion of particles in the box and surface.

Answer 3

Let's be clear on the terminology:

• For a closed system, no mass can enter/leave.
• For an insulated system, no energy can enter/leave.
• For an isolated system, no mass nor energy can enter/leave.

If you really are referring to an insulated or isolated system, then no more math has to be done; you can immediately state that the internal energy $U$ remains unchanged, $\Delta U=0$, by the above definition.

Next, let's be clear on the involved properties (in your case; these definitions will vary):

• $U$ is internal energy.
• $Q$ is heat added into the system.
• $W$ is work done by the system

Note how $Q$ and $W$ do not represent energies, but energy transfers. In your case, $Q=0$ because no energy is being transferred as heat. Sure, the transferred energy becomes thermal energy, but nevertheless it is being transferred as work. So, only $W$ is present,

$$\Delta U=-W.$$

Now, remember that $W$ in this equation, which is the 1st law of thermodynamics adjusted to your case, represents all work. If you have several contributions of work within your system, then they must be added up.

In your scenario, there are two obvious forces: The pulling force and the friction force.

• If you choose your system to be box-and-surface, then only the pulling force $F_\text{pull}$ is external and does work, $$\Delta U=-W=-F_\text{pull} d\neq 0.$$

From Newton's 1st law you can deduce that the friction force equals the pulling force, $F_\mu=F_\text{pull}$, so you are correct in writing

$$\Delta U=-F_\mu d.$$

• But if you include as a part of your system the "thing" that is exerting the pulling force, then you must include the work done on that "thing" as well - which is the same amount of work but with an opposite sign (because energy is being extracted rather than added to this "thing"),

$$\Delta U=-F_\text{pull}d+F_\text{pull}d=0.$$

Only this latter system is isolated, since no energy enters nor leaves. All the energy that is being transferred or converted is already within the system to start with (energy is transferred as work from the "thing" to the box, so overall the system still contains the same amount of energy).