Let the direction of motion down the incline be the x-axis. The net force in this direction determines the acceleration $a$.

$$ \sum F_x = mg\sin\theta - f_k = ma $$

where $f_k$ is the total kinetic friction force.

The friction force is the sum of the friction from the two surfaces of the trough, $f_k = f_{k1} + f_{k2} = \mu_k N_1 + \mu_k N_2$. By symmetry, the normal forces from each surface are equal, $N_1 = N_2 = N$, so the total friction force is $f_k = 2\mu_k N$.

To find the normal force $N$, we analyze the forces in the plane perpendicular to the direction of motion. In this plane, the component of gravity is $mg\cos\theta$, which acts vertically downwards in the cross-sectional view. This force is balanced by the components of the two normal forces.

Since the trough has a 90° angle, each normal force vector makes an angle of 45° with the direction of the $mg\cos\theta$ component. Summing the forces in this direction:

$$ \sum F_\perp = N\cos(45^\circ) + N\cos(45^\circ) - mg\cos\theta = 0 $$ $$ 2N\cos(45^\circ) = mg\cos\theta $$

Solving for N, with $\cos(45^\circ) = 1/\sqrt{2}$:

$$ 2N \left(\frac{1}{\sqrt{2}}\right) = mg\cos\theta \implies N = \frac{mg\cos\theta}{\sqrt{2}} $$

Now, substitute $N$ into the expression for the total friction force:

$$ f_k = 2\mu_k \left(\frac{mg\cos\theta}{\sqrt{2}}\right) = \sqrt{2}\mu_k mg\cos\theta $$

Finally, substitute the friction force back into Newton's second law for the x-direction:

$$ ma = mg\sin\theta - \sqrt{2}\mu_k mg\cos\theta $$

Dividing by the mass $m$ gives the acceleration:

$$ a = g\sin\theta - \sqrt{2}\mu_k g\cos\theta = g(\sin\theta - \sqrt{2}\mu_k \cos\theta) $$