When deriving the Euler-Lagrange equations from the Principle of Stationary Action we start by insisting that the actual path of a particle minimizes the action $S$, defined as $$S=\int_{t_1}^{t_2}\mathcal{L}(q,\dot{q},t),\quad\text{where}\quad\dot{q}=\frac{dq}{dx}.$$ And then we show that a necessary condition for $S$ to be minimized is that $\mathcal{L}$ obeys the Euler-Lagrange equations, $$ \frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\dot{q}}\right)-\frac{\partial\mathcal{L}}{\partial q}=0. $$ My problem is that nowhere in the derivation do we make use of the fact that the Lagrangian is of the form $\mathcal{L}=T-U$, where $T$ is the kinetic energy, and $U$ is the potential energy. So my question is: Is there an explanation for this identification beyond the fact that this what makes the physics work?
My best guess at the answer is that if we instead started out with the Principle of Virtual Work, we eventually get to $$ \frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q}_j}\right)-\frac{\partial T}{\partial q_j}=Q_j. $$ Where $Q_j$ is the generalized force, defined as $$ Q_j\equiv \sum_i\vec{F}_i\cdot\frac{\partial\vec{r}_i}{q_j}, $$ and if we assume that the applied forces on the system are conservative, so that $$ \vec{F}_i=-\vec{\nabla}U(\vec{r}_1,\vec{r}_2,...,\vec{r}_N), $$ and this potential, $U$, doesn't depend on the generalized velocities, we get $$ \frac{d}{dt}\left[\frac{\partial}{\partial \dot{q}_j}(T-U)\right]-\frac{\partial}{\partial q_j}(T-U)=0. $$ Which is exactly the E-L equation, with $\mathcal{L}=T-U$, so we can claim that both principles should give us the same physics, and therefore the Lagrangian we arrived at from the Principle of Stationary Action should also be $\mathcal{L}=T-U$.
Maybe it's just me, but this seems unsatisfactory. Have we simply defined the action so that we have a more compact formulation of the physics we already knew, or can it stand by itself?
