An Apparent Contradiction in the Work—Energy Theorem
Here let me share a confusion I had until yesterday about a relatively embarrassingly basic topic in physics, the idea of work done by a force in classical mechanics. As often strangely occurs, academic ideas come to me when I’m trying to fall asleep that demand attention, and some of these ideas are actually useful and insightful… here is one such example.
The problem which came to my mind while I was half asleep was a situation where we have a guy clinging on to a rope hanging from the ceiling, and we suppose that the guy now climbs up the rope. Let’s analyse the physics, considering the rope + guy as one system.
At first, I was briefly weirded out by how the “reason” why the guy is able to climb up the rope (and hence cause the CM of the system to increase) is actually an internal force to the (rope + guy) system, and internal forces can’t cause acceleration of the system! This was quickly resolved when I realised what was happening was that the contact force of the rope with the ceiling is increased (compared to if the guy were clinging at a fixed position and not climbing), and this is the external force that causes acceleration.
Yet there’s still a problem: what is the work done by this force? The system’s CM is accelerating upwards, and thus gaining in KE, and this is due to the ceiling contact force. By the work—kinetic energy theorem, this should mean that the force did some work, leading to this acceleration, right? But the contact point is not moving… So the work done should be zero. This thoroughly confused me!
For the sake of clarity, here’s the proof of the work—kinetic energy theorem I had in mind; the one that I was always familiar with:
Theorem 1. If a force $\mathbf{F}$ acts on a system with mass $m$, then $\displaystyle\int_{\gamma} \mathbf{F}\cdot\mathrm{d}\mathbf{x} = \Delta\,\mathrm{KE}$, where $\gamma$ is the path taken by the CM of the system.
Proof. By Newton’s Second Law, $$\mathbf{F} = m\mathbf{a} = m\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} = m\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}\mathbf{x}}\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} = m\mathbf{v}\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}\mathbf{x}}.$$ Separating variables, $$\mathbf{F}\mathrm{d}\mathbf{x}=m\mathbf{v}\,\mathrm{d}\mathbf{v}.$$ Integrating both sides yields the conclusion. (We need to be a bit more careful about the vector derivatives and operations, as written this is wrong, but the idea is correct…)
As the proof of this theorem shows, the intuition is that force leads to acceleration which leads to change in KE; the magnitude of the change in KE is equal to the work done. But in the above we seemed to have a scenario where the change in KE is not zero but the work is zero, a contradiction!
What is work?
The resolution to this apparent paradox is that I never understood what work was! More precisely, I had held two different ideas of work in my mind and never realised they were incompatible, and this was leading to the confusion. Let me state these two definitions of work:
The work done by a force on a system is the integral of $\mathbf{F}\cdot\mathrm{d}\mathbf{x}$ across the path of the CM of the system
The work done by a force on a system is the integral of $\mathbf{F}\cdot\mathrm{d}\mathbf{x}$ across the path of the point of action of the force on the system
This is a huge difference! Notice that in my proof of the work—kinetic energy theorem above, I was using the first definition, since when applying N2L to systems of particles, the acceleration is referring to the acceleration of the CM (and hence so are the respective velocities and positions that show up in the derivation). On the other hand, when calculating the work done by the ceiling force in the example, we were using the second definition.
The second definition is more conventional in physics textbooks. But the theorem I proved above does not apply to this definition! We need a different version of the work—kinetic energy theorem that can apply to this definition, which would in fact resolve the paradox.
Theorem 2. If a force $\mathbf{F}$ acts on a system, then $\displaystyle\int_{\gamma} \mathbf{F}\cdot\mathrm{d}\mathbf{x} = \Delta\,\mathrm{KE} + \Delta\,\mathrm{PE}_{\mathrm{internal}}$, where $\gamma$ is the path taken by the action point of the force.
Proof. Since we now are considering a situation where the path taken by the action point could differ from that taken by the CM, then we need to consider each point mass in the system separately to account for this possibility. The KE change of the $i$th particle in this system, except for the one acted upon directly by the external force, is equal to the total work done by all the internal forces: $$\Delta\,\mathrm{KE}_i = \sum_{j}\int\mathbf{F}_{ij}\cdot\mathrm{d}\mathbf{x}_i.$$ Here, $\mathbf{F}_{ij}$ refers to the force acting on the $i$th particle due to the $j$th particle. The particle acted upon directly by the external force will also follow this expression, but with an additional term that corresponds to the work done by the external force. Thus the total change in KE of the system becomes $$\Delta\,\mathrm{KE} = \int\gamma \mathbf{F}\cdot\mathrm{d}\mathbf{x}+ \sum_{i,j}\int\mathbf{F}_{ij}\cdot\mathrm{d}\mathbf{x}_i. $$ In the sum, there occur pairs of the form $$ \int\mathbf{F}_{ij}\cdot\mathrm{d}\mathbf{x}_i+\int\mathbf{F}_{ji}\cdot\mathrm{d}\mathbf{x}_j. $$ By Newton's Third Law we have $\mathbf{F}_{ji}=-\mathbf{F}_{ij}$, and so this can be rewritten as $$\int\mathbf{F}_{ij}\cdot\mathrm{d}\mathbf{x}_i - \int\mathbf{F}_{ij}\cdot\mathrm{d}\mathbf{x}_j = \int\mathbf{F}_{ij}\cdot\mathrm{d}(\mathbf{x}_i-\mathbf{x}_j).$$ If our object is a rigid body undergoing pure translational motion so that $\mathbf{x}_i-\mathbf{x}_j$ would be constant across time, this term vanishes, and we recover a result in the same form as in Theorem 1; however, there is no reason why this should be true in general. The interpretation of this term is that it is negative of the change in potential energy corresponding to the internal forces of the system, so we get the result as claimed.
From this proof, we see that in the second definition of work done, we must account for this extra potential energy term in the calculation to get correct results. This, then, resolves the apparent contradiction in our rope problem: the work done is zero, and the KE increases, but it’s okay because the internal PE of the system decreases.
