(i) The loss in gravitational potential energy (PE) is given by \(mgh\), where \(m = 15,000\) kg, \(g = 9.8\) m/s², and \(h = 18\) m. Thus,

\(\text{Loss in PE} = 15,000 \times 9.8 \times 18 = 2.7 \times 10^6 \text{ J}\)

(ii) The work done by the driving force (WD) is the difference between the loss in potential energy and the work done by resistive forces. Therefore,

\(\text{WD} = 2.7 \times 10^6 - 4.8 \times 10^6 = 2.1 \times 10^6 \text{ J}\)

(iii) The change in kinetic energy (KE) as the lorry speeds up from 14 m/s to 16 m/s is given by

\(\Delta KE = \frac{1}{2} \times 15,000 \times (16^2 - 14^2)\)

\(= \frac{1}{2} \times 15,000 \times (256 - 196) = 450,000 \text{ J}\)

The work done by the driving force (WD by DF) is the sum of the gain in kinetic energy and the work done against resistance:

\(\text{WD by DF} = 450,000 + 1600 \times 2500\)

\(= 450,000 + 4,000,000 = 4.45 \times 10^6 \text{ J}\)