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Let's consider a setup consisting of a table with friction, and a block on top of it. Suppose we drag the block across the table with a constant speed. The applied force $f_{app}$ acting through a distance $d$ does a work $f_{app}d$. The frictional force $\mu N$ is equal to $f_{app}$ since there is no acceleration. So the total work done on the block by these external forces will be $\Sigma W = W_{app} + W_{friction} = F_{app}d - f_{Kinetic Friction}d = 0$

According to law of conservation of energy, if there is an energy change in the system , it is because the energy is being transferred across the system boundary by a transfer mechanism (work, heat, mechanical waves, matter transfer, electromagnetic radiation etc.)
So, $\Delta E_{system} = \Sigma T$, where $\Delta E_{system} = \Delta K + \Delta U + \Delta E_{internal}$ and $\Sigma T$ is an energy transfer mechanism.

In our case, we have $ W_{app} + W_{friction} = \Delta K + \Delta E_{internal} $, $\Delta K = 0$ as there is no change in speed, then $ W_{app} + W_{friction} = \Delta E_{internal} $ , but as $ W_{app} + W_{friction} = 0 \Longrightarrow \Delta E_{internal} = 0$ So that means there is no increase in the internal energy, but clearly the block heats up. Can anyone explain to me what's happening here?

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In order to write the expression

$$W_{app}+W_{friction}=\Delta K + \Delta E_{internal}$$

you had to assume $\Delta E_{system}=W_{app}+W_{friction}$. But in that statement, you are assuming a priori there is no change in energy in the system due to heat.

I think a more careful analysis would be to start with $\Delta E_{system}=\Delta K+ \Delta U + \Delta E_{internal}$, use the work-energy theorem to write $W_{app}+W_{friction}=\Delta K$ and plug that in. That gives you (since $\Delta K=0$, as you've noticed),

$$\Delta E_{system}=\Delta U + \Delta E_{internal}$$

and if there is no change in potential energy, the energy change in the system is equal to the internal energy change, which would be due to the thermal contact between the floor and the block.

EDIT: "Which form of the conservation of energy equation should we use"? I would argue that there is only one conservation of energy equation, which is

$$\Delta E_{system}=\Sigma E_{interactions}$$ and the challenge is to determine the source and model for each interaction. That is, you have to specify the interactions (gravational, E/M, friction, heat, photon flux, whatever). Once you specify the possible interactions, you're not even done - you also have to specify how the interactions transfer energy. Here are some possibilies:

  • Nonconservative force, $\Delta E=\vec{F}\cdot d\vec{x}$.
  • Conservative force, $\Delta E=-\Delta U$.
  • Change in volume, $\Delta E=PdV$
  • Heat transfer, $\Delta E=c\Delta T$
  • Motion, $\Delta E=\frac{1}{2}mv^2$.

This is all rather schematic, but I hope it gets my point across. You really can't use conservation of energy to look for "missing energy" unless you specify beforehand "all the missing energy will be found in "

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  • $\begingroup$ So you are implying what I am missing is an expression for heat transfer. The correct expression is $\Delta E_{system} = \Sigma W + Q$ that accounts for the change in the internal energy? $\endgroup$ – Nuuh Sep 18 '17 at 17:25
  • $\begingroup$ That is also a statement of energy conservation, but with $W$ standing more for "thermodynamic work" rather than "work due to a force". Of course, all these statements mean the exact same thing about energy transfer, you just have to be sure not to assume a particular source of energy is zero if it's in your problem. $\endgroup$ – levitopher Sep 18 '17 at 17:31
  • $\begingroup$ Why does $W$ stand for thermodynamic work? Shouldn't it stand for the work done by the forces, as I am using a more general equation for energy conservation which accounts for all types of energy transfers? I thought $Q$ stood for heat transfer which explained why there is a change in the internal energy. $\endgroup$ – Nuuh Sep 18 '17 at 17:38
  • $\begingroup$ @Nuuh, If you are doing some work on the block and there is no change in the block's KE, then it is this work done by you that emerges out as the heat, which turns out to be the work by frictional force. Thermodynamic work is the one that changes the volume of the system. Since we consider bodies to be perfectly rigid in classical mechanics, there is no thermodynamic work. Internal energy remains the same. $\endgroup$ – Mitchell Sep 18 '17 at 17:45
  • $\begingroup$ If the work done by me emerges as internal energy change, shouldn't $\Delta E_{system} = W_{app} + W_{frictional} = \Delta K + \Delta U + \Delta E_{int}$ account for $\Delta_{int}$, the change in internal energy ? The above expression is telling me $\Delta_{int} = 0$ which says there will be no change internal energy. $\endgroup$ – Nuuh Sep 18 '17 at 18:06

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