= Solution
Let $K=\rho_0|\mathbf u|^2/2$ be the <kinetic energy> per unit volume. Dot the momentum equation with $\rho_0\mathbf u$. The <Coriolis acceleration> does no work because $\mathbf u\cdot(\mathbf e_z\times\mathbf u)=0$. <Incompressibility> turns the <advection> and <pressure> terms into divergences, yielding
$$
\boxed{\partial_tK+\nabla\cdot\bigl[(K+P)\mathbf u\bigr]=\rho_0\Omega^2(3x-2qz)u_x.}
$$
Thus $\mathbf F=(K+P)\mathbf u$ is the <energy flux> and $S=\rho_0\Omega^2(3x-2qz)u_x$ is the source in the requested kinetic-energy form.
To decide about <conservation of energy>, absorb the ordinary radial <shearing-sheet tidal potential> $\Phi_t=-3\Omega^2x^2/2$ into the density. Its material derivative cancels the $3x$ work term:
$$
\boxed{\partial_t(K+\rho_0\Phi_t)+\nabla\cdot\bigl[(K+P+\rho_0\Phi_t)\mathbf u\bigr]=-2\rho_0\Omega^2qz\,u_x.}
$$
For $q=0$ this is the usual conservative <Jacobi energy in a shearing sheet>, subject to vanishing boundary energy flux. For $q\ne0$, the extra body force has nonzero <curl>, $\nabla\times[-2\Omega^2qz\mathbf e_x]=-2\Omega^2q\mathbf e_y$, so no scalar potential can absorb it. A potential $2\Omega^2qxz$ would necessarily introduce a vertical force $-2\Omega^2qx\mathbf e_z$, absent from the model. Therefore \b[this vertical-shear model does not generically conserve the standard total mechanical energy]: its maintained shear/irradiation background can supply or remove energy. This is the <energy balance of the incompressible vertical-shear model>.
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