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!

  • $\begingroup$ I posted a similar question some months ago. There is an article that adresses this issue: "The Work-Energy Theorem and the First Law of Thermodynamics" by José María Mínguez. $\endgroup$
    – Rejalgar
    Jul 13, 2023 at 13:13

2 Answers 2


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.

  • $\begingroup$ (1/n) Thanks for your response! Can you maybe provide a link so that I can read up further about the work-energy theorem as it pertains to multi-particle systems? Because every link I find as well as all my textbooks speak solely about it in relation to single particles which I think is the source of my confusion. Let me also try explain why I think the 1st law is a more general form of the W-E theorem. Suppose we have a mass m resting on the earth and we assume negligible air resistance. Now let the mass m and the earth be the system. $\endgroup$ Feb 13, 2021 at 4:53
  • $\begingroup$ (2/n) If I (the surroundings) apply a force ($F_{a}=mg+F'$) on the the mass for 1 meter so that the mass rises 1 meter, the net work done on the system is $W_{net}=mg+F'$. But clearly this net work is not equal to $\Delta K_{sys}$. So we have that $W_{net} \neq \Delta K $ which goes against the W-E theorem. But according to the first law we have that $\Delta U_{sys} =Q+W$ and in our case Q=0 so that we get $\Delta U_{sys} =mg+F'$. Now for our system $\Delta U_{sys} = \Delta K_{sys} + \Delta V_{sys}$. But because the earth is large we get that $\Delta K_{sys}=\Delta K_{m}$. $\endgroup$ Feb 13, 2021 at 5:03
  • $\begingroup$ (3/n) Now we can calculate $\Delta K_m$ by applying the W-E thm to the single particle system m because the W-E thm does apply to single particles. Doing that we get $W_{net,m}=F'$ because the work done by gravity must be subtracted off. And so $\Delta K_m = W_{net,m}=F' = \Delta K_{sys}$. Thus we finally get that $\Delta U_{sys}=mg+F'=F'+\Delta V_{sys} \Rightarrow \Delta V_{sys}=mg $ and everything works out. This example seems to show that the more special W-E thm is too narrow to solve for $\Delta V_{sys}$ but the 1st law being more general was easily used to help solve for $\Delta V_{s}$ $\endgroup$ Feb 13, 2021 at 5:11
  • $\begingroup$ Work-energy theorem is badly covered in modern textbooks/on the Internet, you have to go for the older/original sources. See e.g. Synge J.L., Griffith B.A.: Principles of Mechanics (1949), sec 5.2, part "The principle of energy." Djvu file for this book is available on the Internet. $\endgroup$ Feb 13, 2021 at 13:43
  • $\begingroup$ Your calculation does not make much sense to me. The work-energy theorem does not talk about potential energy, only about kinetic energy. For your example, if the body is raised very slowly, and external force balances the gravity force, the work-energy theorem predicts that zero work means zero change in kinetic energy, which is correct. $\endgroup$ Feb 13, 2021 at 13:46

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

$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad (18) $$

  • $\begingroup$ So is it correct to state that the work-energy theorem only applies to single-body rigid systems? And then is it also correct to state that the first law of thermodynamics is a more general form of the work-energy theorem which we can use for multi-particle deformable systems to determine energy changes due to work? $\endgroup$ Feb 12, 2021 at 14:19
  • $\begingroup$ @SalahTheGoat There is no inherent restriction to single body mechanics. For any body subject to multiple forces the resultant motion is determined by the vector sum of the forces involved. It is assumed (and no counter-evidence is known) that pressure of a gas can be understood in terms of the mechanics of collisions. In terms of statistical mechanics the atoms that comprise matter have no stable way of storing energy. The name 'thermodynamics' is used for the discipline where physics is not stated in terms of motions of atoms, so then one ends up formulating a concept of internal energy. $\endgroup$
    – Cleonis
    Feb 12, 2021 at 14:56

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