(i) The forces acting on the box are the gravitational component down the slope \(F = 50g \sin 10^{\circ}\) and the normal reaction \(R = 50g \cos 10^{\circ}\). For the box to remain at rest, the frictional force must be at least equal to the component of gravity down the slope. Therefore, the coefficient of friction \(\mu\) must satisfy:

\(\mu \geq \frac{F}{R} = \frac{50g \sin 10^{\circ}}{50g \cos 10^{\circ}} = \tan 10^{\circ}\)

Calculating \(\tan 10^{\circ}\), we find \(\mu \geq 0.176\).

(ii) The work done by the girl is \(50 \times 5\). The potential energy loss is \(50g \times 10 \sin 10^{\circ}\). The work done against friction over 10 m is \(0.19 \times 50g \cos 10^{\circ} \times 10\). Using the energy method:

\(50 \times 5 + 50g \times 10 \sin 10^{\circ} - 0.19 \times 50g \cos 10^{\circ} \times 10 = 0.5 \times 50v^2\)

Solving for \(v\), we find \(v = 2.70 \text{ m/s}\).

(iii) On the 20° incline, the net force is \(50g \sin 20^{\circ} - 0.19 \times 50g \cos 20^{\circ}\). Using Newton's second law:

\(50g \sin 20^{\circ} - 0.19 \times 50g \cos 20^{\circ} = 50a\)

Solving for \(a\), we find \(a = 1.63 \text{ m/s}^2\).