(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:

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

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

(i) Using conservation of energy, the potential energy at A is converted to kinetic energy at E. The equation is:

\(mgh = \frac{1}{2} m (6)^2\)

Solving for \(h\):

\(gh = \frac{1}{2} (6)^2\)

\(h = \frac{18}{g}\)

Assuming \(g = 9.8 \text{ m/s}^2\),

\(h = \frac{18}{9.8} \approx 1.8 \text{ m}\)

(ii) The maximum speed occurs at the lowest point C. Using energy conservation:

\(\frac{1}{2} mv^2 = mg(1.8 + 0.65)\)

\(v^2 = 2g(2.45)\)

\(v = \sqrt{2 \times 9.8 \times 2.45} \approx 7 \text{ m/s}\)

Given the power of the engine is constant at 15,000 W, we use the formula for power:

\(P = Fv\)

At point A, the force \(F_A = 750\) N, so:

\(15,000 = 750v_A\)

\(v_A = \frac{15,000}{750} = 20 \text{ m/s}\)

At point B, the force \(F_B = 500\) N, so:

\(15,000 = 500v_B\)

\(v_B = \frac{15,000}{500} = 30 \text{ m/s}\)

The increase in kinetic energy is given by:

\(\Delta KE = \frac{1}{2} m (v_B^2 - v_A^2)\)

\(\Delta KE = \frac{1}{2} \times 1000 \times (30^2 - 20^2)\)

\(\Delta KE = 500 \times (900 - 400)\)

\(\Delta KE = 500 \times 500 = 250,000 \text{ J}\)

Thus, the increase in kinetic energy is 250,000 J or 250 kJ.

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

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

where \(m = 80 \text{ kg}\), \(v = 9 \text{ m/s}\), and \(u = 0 \text{ m/s}\).

\(\Delta KE = \frac{1}{2} \times 80 \times (9^2 - 0^2) = 3240 \text{ J}\)

The loss in potential energy (PE) is given by:

\(\Delta PE = mgh\)

where \(h = 250 \times \sin(2.6^{\circ})\).

\(\Delta PE = 80 \times 9.8 \times 250 \times \sin(2.6^{\circ}) = 9070 \text{ J}\)

The work done against resistance is:

\(\text{Work done} = \Delta PE - \Delta KE = 9070 - 3240 = 5830 \text{ J}\)

(ii) Using the work-energy principle, the work done is also equal to the loss in kinetic energy along BC:

\(5830 = \frac{1}{2} \times 80 \times (9^2 - 5^2)\)

Solving for \(d\):

\(5830 = 80 \times 9.8 \times d\)

\(d = \frac{5830}{80 \times 9.8} = 96.0 \text{ m}\)

(iii) The driving force is given by:

\(\text{Driving force} = \frac{425}{5} = 85 \text{ N}\)

Using Newton's second law:

\(85 - 23.3 = 80a\)

\(a = \frac{85 - 23.3}{80} = 0.771 \text{ m/s}^2\)

The total work done by the pulling force is the sum of the work done against resistance and the gain in potential energy:

\(\text{Total work done} = 900 \text{ J} + 2100 \text{ J} = 3000 \text{ J}\)

The work done by the force \(F\) is given by:

\(\text{Work done} = F \times 100 \times \cos(15^{\circ})\)

Equating the two expressions for work done:

\(3000 = F \times 100 \times \cos(15^{\circ})\)

Solving for \(F\):

\(F = \frac{3000}{100 \times \cos(15^{\circ})}\)

Calculating \(F\):

\(F \approx 31.1 \text{ N}\)