Step 1: Recall the formula for power.
Power is the rate at which a force does work. When a force $F$ acts on an object moving at velocity $v$ and the angle between them is $\theta$, the power is: \[ P = Fv\cos\theta \]
Step 2: Identify the given quantities.
Force: $F = 500$ N. Speed: $v = 4$ m/s. Angle of force above horizontal: $\theta = 30^\circ$. The velocity is horizontal, so $\theta$ is the angle between the force and the velocity direction.
Step 3: Substitute into the power formula.
\[ P = Fv\cos\theta = 500 \times 4 \times \cos 30^\circ \] \[ P = 2000 \times \frac{\sqrt{3}}{2} = 1000\sqrt{3} \text{ W} \]
Step 4: Calculate the numerical value.
\[ P = 1000\sqrt{3} \approx 1000 \times 1.732 = 1732 \text{ W} \]
Step 5: Note that the block moves at constant speed.
Since the block moves at constant speed, the net work done is zero, but the applied force still does positive work at the rate calculated. The friction force does negative work that cancels it out. The question asks for the rate of work done by the applied force only.
Step 6: State the final answer.
The rate at which the applied force does work on the block is $1000\sqrt{3} \approx 1732$ W. \[ \boxed{1732 \text{ W}} \]