(i) The loss of kinetic energy (KE) is calculated using the formula:

\(\text{KE loss} = \frac{1}{2} m (u^2 - v^2)\)

where \(m = 20 \text{ kg}\), \(u = 2.5 \text{ m/s}\), and \(v = 1.5 \text{ m/s}\).

\(\text{KE loss} = \frac{1}{2} \times 20 \times (2.5^2 - 1.5^2) = 40 \text{ J}\)

The gain in potential energy (PE) is calculated using:

\(\text{PE gain} = mgL \sin \alpha\)

where \(g = 9.8 \text{ m/s}^2\), \(L = 10 \text{ m}\), and \(\alpha = 4.5^{\circ}\).

\(\text{PE gain} = 20 \times 9.8 \times 10 \times \sin 4.5^{\circ} = 157 \text{ J}\)

(ii) The work done by the pulling force is given by:

\(\text{Work done} = \text{PE gain} - \text{KE loss} + \text{work against resistance}\)

\(\text{Work done} = 157 - 40 + 50 = 167 \text{ J}\)

(iii) The magnitude of the pulling force \(F\) is found using:

\(\text{Work done} = F \cdot L \cdot \cos 15^{\circ}\)

\(167 = F \times 10 \times \cos 15^{\circ}\)

\(F = \frac{167}{10 \times \cos 15^{\circ}} = 17.3 \text{ N}\)