(i) Using the equation of motion, \(v = u + at\), where \(v = 2 \text{ m/s}\), \(u = 0 \text{ m/s}\), and \(t = 5 \text{ s}\), we have:

\(2 = 0 + 5a\)

\(a = 0.4 \text{ m/s}^2\)

Applying Newton's second law along the plane:

\(0.1g \sin 20^{\circ} - F = 0.1 \times 0.4\)

\(F = 0.1g \sin 20^{\circ} - 0.04\)

\(F = 0.302 \text{ N}\)

(ii) The normal reaction \(R\) is given by:

\(R = 0.1g \cos 20^{\circ} \approx 0.9397 \text{ N}\)

The coefficient of friction \(\mu\) is:

\(\mu = \frac{F}{R} = \frac{0.302}{0.9397} \approx 0.321\)