First, calculate the work done by the 120 N force:

\(\text{Work done} = 120 \times 5 \times \cos 20^{\circ} = 563.81557 \ldots\)

Next, calculate the change in potential energy:

\(\text{PE change} = 10g \times 5 \times \sin 30^{\circ} = 250\)

Apply the work-energy principle:

\(120 \times 5 \cos 20^{\circ} - 10g \times 5 \sin 30^{\circ} - 200 = \frac{1}{2} \times 10 \times v^2\)

\(563.815 \ldots - 250 - 200 = 5v^2\)

Solve for \(v\):

\(v^2 = \frac{113.815 \ldots}{5}\)

\(v = \sqrt{22.763 \ldots} \approx 4.77 \text{ ms}^{-1}\)