(a) The change in potential energy (PE) is given by \(\Delta PE = mgh\), where \(h = s \sin \theta\). Here, \(s = 30 \times 20 = 600 \text{ m}\) and \(\sin \theta = 0.12\). Thus, \(h = 600 \times 0.12 = 72 \text{ m}\).

\(\Delta PE = 1600 \times 9.8 \times 72 = 1152000 \text{ J}\).

(b) The total work done \(W = 1960 \times 1000 = 1960000 \text{ J}\). The work done against resistance \(W_D = W - \Delta PE = 1960000 - 1152000 = 808000 \text{ J}\).

The force resisting motion \(R = \frac{W_D}{s} = \frac{808000}{600} = 1350 \text{ N}\).

(c) The power developed \(P = \frac{W}{t} = \frac{1960000}{30} = 65333.33 \text{ W} = 65.3 \text{ kW}\).

(d) The new power is \(0.85 \times 65333.33 = 55533.33 \text{ W}\). Using \(P = Fv\), the driving force \(DF = \frac{55533.33}{20} = 2776.67 \text{ N}\).

Using Newton's second law, \(DF - R - mg \sin \theta = ma\).

\(2776.67 - 1350 - 1600 \times 9.8 \times 0.12 = 1600a\).

\(a = \frac{-490}{1600} = -0.306 \text{ m s}^{-2}\).