(i) The driving force (DF) required to move the lorry up the hill is given by the sum of the resistance to motion and the component of the weight of the lorry acting down the slope. The component of the weight down the slope is \(12,000 \times g \times 0.08\), where \(g\) is the acceleration due to gravity (approximately 9.8 m s^-2). Therefore,

\(\text{DF} = 1500 + 12,000 \times 9.8 \times 0.08 = 11,100 \text{ N}\)

The power \(P\) of the engine is given by \(P = \text{DF} \times v\), where \(v = 5 \text{ m s}^{-1}\). Thus,

\(P = 11,100 \times 5 = 55,500 \text{ W} = 55.5 \text{ kW}\)

(ii) Given that the resistance to motion is \(kv^2\) and at \(v = 5 \text{ m s}^{-1}\), the resistance is 1500 N, we have:

\(k \times 5^2 = 1500\)

\(25k = 1500\)

\(k = 60\)

(iii) On a straight level road, the driving force is equal to the resistance, which is \(60v^2\). The power is still 55.5 kW, so:

\(55,500 = 60v^2 \times v = 60v^3\)

\(v^3 = \frac{55,500}{60}\)

\(v^3 = 925\)

\(v = 9.74 \text{ m s}^{-1}\)