A 10 kg box is on a horizontal surface. A force of 200 N is applied at an angle of 30 degrees below the horizontal. The box moves at a constant speed. Calculate the normal force and the friction force

Since the box moves at a constant speed, its acceleration is zero. This means the net force acting on the box in both the horizontal and vertical directions is zero.

$$\sum F_x = 0$$ $$\sum F_y = 0$$

The forces acting vertically are the normal force ($N$) upwards, the weight ($mg$) downwards, and the vertical component of the applied force ($F \sin 30^\circ$) downwards.

$$N - mg - F \sin 30^\circ = 0$$ $$N = mg + F \sin 30^\circ$$

The forces acting horizontally are the horizontal component of the applied force ($F \cos 30^\circ$) in the direction of motion and the kinetic friction force ($f_k = \mu N$) opposite to the direction of motion.

$$F \cos 30^\circ - f_k = 0$$ $$F \cos 30^\circ - \mu N = 0$$

We are given $F = 200 \, \text{N}$, $\mu = 0.4$, $\theta = 30^\circ$, and $g = 10 \, \text{m/s}^2$.

$$F \cos 30^\circ - \mu (mg + F \sin 30^\circ) = 0$$

Substitute the known values:

$$200 \cos 30^\circ - 0.4 (m \cdot 10 + 200 \sin 30^\circ) = 0$$

Using $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$:

$$200 \left(\frac{\sqrt{3}}{2}\right) - 0.4 \left(10m + 200 \cdot \frac{1}{2}\right) = 0$$ $$100\sqrt{3} - 0.4 (10m + 100) = 0$$ $$100\sqrt{3} - 4m - 40 = 0$$

Rearrange to solve for $m$:

$$4m = 100\sqrt{3} - 40$$ $$m = \frac{100\sqrt{3} - 40}{4}$$ $$m = 25\sqrt{3} - 10$$

Calculate the numerical value:

$$m \approx 25(1.732) - 10$$ $$m \approx 43.3 - 10$$ $$m \approx 33.3 \, \text{kg}$$

The value of $m$ is $\boxed{\mathbf{33.3 \, \text{kg}}}$.