(i)(a) For the smooth plane, apply Newton's second law parallel to the plane:

\(200 - 30g \sin 20^{\circ} = 30a\)

\(a = \frac{200 - 30 \times 9.8 \times \sin 20^{\circ}}{30} = 3.25 \text{ ms}^{-2}\)

(i)(b) Using the work-energy principle, the change in kinetic energy \(\Delta KE\) is:

\(\Delta KE = \frac{1}{2} \times 30 \times v^2\)

Using \(v^2 = 2as\), where \(s = 12 \text{ m}\):

\(v^2 = 2 \times 3.25 \times 12 = 77.9\)

\(\Delta KE = \frac{1}{2} \times 30 \times 77.9 = 1170 \text{ J}\)

(ii)(a) For the rough plane, resolve forces perpendicular to the plane to find the normal force \(N\):

\(N = 30g \cos 20^{\circ}\)

The frictional force \(F = \mu N = 0.12 \times 30g \cos 20^{\circ}\)

Apply Newton's second law parallel to the plane:

\(200 - 30g \sin 20^{\circ} - F = 30a\)

\(a = \frac{200 - 30 \times 9.8 \times \sin 20^{\circ} - 0.12 \times 30 \times 9.8 \times \cos 20^{\circ}}{30} = 2.12 \text{ ms}^{-2}\)

(ii)(b) When the force acts at 10° above the line of greatest slope, resolve forces perpendicular to the plane:

\(N + 200 \sin 10^{\circ} = 30g \cos 20^{\circ}\)

\(N = 30g \cos 20^{\circ} - 200 \sin 10^{\circ}\)

The frictional force \(F = 0.12 \times N\)

Apply Newton's second law parallel to the plane:

\(200 \cos 10^{\circ} - 30g \sin 20^{\circ} - F = 30a\)

\(a = \frac{200 \cos 10^{\circ} - 30 \times 9.8 \times \sin 20^{\circ} - 0.12 \times (30g \cos 20^{\circ} - 200 \sin 10^{\circ})}{30} = 2.16 \text{ ms}^{-2}\)