How does the work-energy theorem relate to the first law of thermodynamics? The work energy theorem states that the net work on a particle is equal to the change in the kinetic energy of the particle:
$$W_{net}=\Delta K $$
My first question is whether this formula (the work-energy theorem) only applies to single-body rigid systems, that is, whether it only applies to a single particle that cannot be deformed? It is intuitive to me that it should only apply to single-body rigid systems because if the system is not a single-body rigid system, then the system will have the ability to store potential energy and the net work done on the system could lead to an increase in the systems potential energy as opposed to only changing the systems kinetic energy.
My second question is based on the assumption that the work-energy theorem is in fact only true for single-particle rigid systems. If this assumption is true, then is the first law of thermodynamics ( $\Delta U = Q + W $) simply a more general form of the work-energy theorem that can be applied to any system regardless of the amount of particles in the system and regardless of whether they are deformable?
If the first law is a more general form of the work-energy theorem then is my thinking below correct? Suppose we have a system of two particles connected by a spring. Now suppose I apply an external force ( $F_a$ ) on the system over a distance $\Delta x$ (say I give the first particle a push inward toward the second particle compressing the spring but also causing the systems center of mass to move slightly). Since this system is not a single-particle rigid system, we cannot say that $W_{net} = \Delta K$ because the spring compression means the work has lead to an increase in potential energy as well. But $ \Delta U = Q+W $does still apply. And in this case, we clearly have $Q=0$ and also that $W_{net}=F_a \Delta x $
so that we get $\Delta U=F_a \Delta X $ .  Now for our system we have that $\Delta U= \Delta V +\Delta K$ . But we do not posses enough information to calculate how much of the work went into increasing the potential energy or went into increasing kinetic energy. All we know for sure is that the change in internal energy equal to the net work done on the system. Is this all correct or am I mistaken?
Any help on this issue would be most appreciated!
 A: The work energy theorem (in classical mechanics) is more general than you describe, it applies to any system of particles, even deformable bodies that interact via instantaneous forces, provided work of all forces is accounted for, including work of internal forces. We can formulate the theorem thus:
Sum total of work of all forces, external and internal, on a many particle system, equals change in its kinetic energy.
If we do not account for work of internal forces, like when the body is deformable but we ignore work of force due to one part of the body on the other part, then the work-energy theorem does not hold.
1st law of thermodynamics is similar to work-energy theorem, but it is not "more general". This is because 1) it is about thermodynamic systems, not mechanical systems; 2) it refers to concept of internal energy, which does not include mechanical kinetic energy but does include mechanical potential energy; 3) it says something very different about the internal energy, than the work-energy theorem says about kinetic energy. We can formulate 1st law thus:
For any body there exists a internal energy function $U$ of equilibrium state $X$ which can change via heat transfer $\Delta Q$ or work transfer $\Delta W$ with the surrounding bodies; since the effect of heat transfer can be equally achieved by some equivalent amount of work transfer, we use the same units for both work and heat:
$$
\Delta U = \Delta W + \Delta Q.
$$
Notice how the law introduces a new quantity, internal energy. This is very different from the work-energy theorem, which only says how much kinetic energy increases as a result of work.
A: You mention spring compression.
The procedure to calibrate a spring is as follows: you compress the spring, and over the entire travel of the spring you measure the force that it exerts. That generates a force-displacement profile. To find the potential energy stored in a spring for any value of the displacement you integrate the force over distance, using the profile you measured. That procedure is the way to arrive at a mathematical expression for the potential energy stored in a spring.

The work-energy theorem gives a relation between force acting over distance and change of kinetic energy.
$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2  \qquad (1) $$
The work-energy theorem is - as the name says - a theorem, not a definition.



For completeness the derivation of the work-energy theorem
In the course of the derivation the following two relations will be used:
$$ ds = v \ dt  \qquad (11)  $$
$$ dv = \frac{dv}{dt}dt  \qquad (12)   $$
The integral for acceleration from a starting point $s_0$ to a final point $s$
$$ \int_{s_0}^s a \ ds \qquad (13)  $$
Use (11) to change the differential from ds to dt. Since the differential is changed the limits change accordingly.
$$ \int_{t_0}^t a \ v \ dt \qquad (14) $$
Rearrange the order, and write the acceleration $a$ as $\tfrac{dv}{dt}$
$$ \int_{t_0}^t v \ \frac{dv}{dt} \ dt  \qquad (15) $$
Use (12) for a second change of differential, again the limits change accordingly.
$$ \int_{v_0}^v v \ dv  \qquad (16)  $$
Putting everything together:
$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2  \qquad (17) $$
Combining with $F=ma$ gives the work-energy theorem
$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2  \qquad (18) $$
