(i) First, calculate the normal reaction force, \(R\), using:

\(R = 0.8g \cos 10^{\circ} \approx 7.88 \text{ N}\)

The frictional force, \(F\), is given by:

\(F = \mu R = 0.4 \times 0.8g \cos 10^{\circ} \approx 3.15 \text{ N}\)

Using Newton's second law along the plane:

\(-0.8g \sin 10^{\circ} - F = 0.8a\)

\(-0.8g \sin 10^{\circ} - 3.15 = 0.8a\)

Solving for \(a\):

\(a = \frac{-0.8g \sin 10^{\circ} - 3.15}{0.8} \approx -5.68 \text{ m s}^{-2}\)

(ii) Using the equation of motion \(v^2 = u^2 + 2as\) with final velocity \(v = 0\), initial velocity \(u = 12 \text{ m s}^{-1}\), and acceleration \(a = -5.68 \text{ m s}^{-2}\):

\(0 = 12^2 + 2(-5.68)s\)

\(s = \frac{144}{2 \times 5.68} \approx 12.7 \text{ m}\)