Consider an individual charge carrier in the wire. I cite a portion of one of my previous answers:What is the mechanism by which magnetic fields do work?.
As the diagram below shows, the initial magnetic force will be
upwards, perpendicular to the current. No work is being done. Since
the charge carriers are constrained to move in the wire, they will
exert a force (not magnetic in nature) on the wire. Hence the wire
will start moving upwards.
Now the instant the wire starts moving upwards, the charges will also
gain a velocity component $\textbf{v}^\perp$ perpendicular to their
original motion. This component will produce a horizontal component
$\textbf{F}_{mag}^\perp$ of the magnetic force opposing the flow of
current. In the absence of a source then, the currents will decrease,
and the work done by the magnetic field in doing so will precisely
balance out the work done by the horizontal component in raising the
ring. Hence the magnetic field will have done no work, and of course
we see this visually because the total force $\textbf{F}_{mag}$ is
indeed perpendicular to the total velocity $\textbf{v}$.
Take the integral of $\textbf{F}^\perp_{mag}$ along the length of the rod and you get your motional emf. Thus, the motional emf is really just a component of the Lorentz force, which just so happens to have the form:
\begin{equation}
\varepsilon = -\frac{d\Phi}{dt}
\end{equation}
which is the same as for a Faraday emf (which is instead due to a changing magnetic field). This is not a coincidence (of course), and is what led Einstein to his formulation of special relativity.
This explanation can be easily extended to a rod moving on two rails.
EDIT: i don't know if this is useful in your context but as the rod and the two rails form a circuit one would also need to take into account the self-inductance. When the current in the circuit changes the magnetic field produced by it will change. This will induce a back-emf opposing the change in current.
