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

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

where \(m = 50 \text{ kg}\), \(u = 7 \text{ m/s}\), and \(v = 3 \text{ m/s}\).

\(\text{Loss in KE} = \frac{1}{2} \times 50 \times (3^2 - 7^2) = 1000 \text{ J}\)

(ii) The gain in potential energy (PE) is given by:

\(\text{Gain in PE} = mgh\)

where \(h = 15 \text{ m}\) and \(g = 10 \text{ m/s}^2\).

\(\text{Gain in PE} = 50 \times 10 \times 15 = 7500 \text{ J}\)

(iii) The work done (WD) by the pulling force is calculated by considering the work done against resistance and the change in energy:

\(\text{WD by resistance} = 7.5 \times 200 = 1500 \text{ J}\)

\(\text{WD by pulling force} = \text{Gain in PE} + \text{Loss in KE} + \text{WD by resistance}\)

\(\text{WD by pulling force} = 7500 + 1500 - 1000 = 8000 \text{ J}\)

(iv) To find \(\alpha\), use the formula for work done by the force:

\(\text{WD} = F \cdot d \cdot \cos \alpha\)

where \(F = 45 \text{ N}\) and \(d = 200 \text{ m}\).

\(8000 = 45 \times 200 \times \cos \alpha\)

\(\cos \alpha = \frac{8000}{45 \times 200}\)

\(\alpha = \cos^{-1} \left( \frac{8000}{9000} \right) \approx 27.3 \degree\)