The box moves at a constant velocity, so the net force is zero. We use a coordinate system with the x-axis parallel to the incline (upwards) and the y-axis perpendicular to it. Sum of forces perpendicular to the incline: $N - mg \cos\theta - F \sin\theta = 0 \implies N = mg \cos\theta + F \sin\theta$. Sum of forces parallel to the incline: $F \cos\theta - mg \sin\theta - f = 0$. The friction force is $f = \mu N = \mu(mg \cos\theta + F \sin\theta)$. Substitute $f$ into the parallel force equation:

$F \cos\theta - mg \sin\theta - \mu(mg \cos\theta + F \sin\theta) = 0$.

Solving for $F$:

$$F(\cos\theta - \mu \sin\theta) = mg(\sin\theta + \mu \cos\theta) \implies F = mg \frac{\sin\theta + \mu \cos\theta}{\cos\theta - \mu \sin\theta}$$

Substitute values ($g = 9.8$ m/s$^2$):

$F = 50 \cdot 9.8 \frac{\sin(30^\circ) + 0.20 \cos(30^\circ)}{\cos(30^\circ) - 0.20 \sin(30^\circ)} \approx 430.5$ N.

The friction force is $f = F \cos\theta - mg \sin\theta = 430.5 \cos(30^\circ) - 50 \cdot 9.8 \sin(30^\circ) \approx 127.9$ N.

The work done by each force is calculated as follows: Work by pushing force $F$: The angle between $F$ and displacement is $\theta$.

$W_F = F s \cos\theta = 430.5 \text{ N} \cdot 6.0 \text{ m} \cdot \cos(30^\circ) \approx 2.2 \times 10^3$ J.

Work by friction $f$: Friction opposes motion.

$W_f = -fs = -127.9 \text{ N} \cdot 6.0 \text{ m} \approx -7.7 \times 10^2$ J.

Work by normal force $N$: $N$ is perpendicular to displacement.

$W_N = 0$ J.

Work by gravity $G$: Gravity acts vertically, displacement is up the incline.

$W_G = -mgh = -mgs \sin\theta = -50 \text{ kg} \cdot 9.8 \text{ m/s}^2 \cdot 6.0 \text{ m} \cdot \sin(30^\circ) = -1.5 \times 10^3$ J.

Work by net force: By the work-energy theorem, since velocity is constant, $\Delta E_k = 0$.

$W_{net} = \Delta E_k = 0$ J.