(i) The horizontal component of the tension is given by:

\(T \cos(20^{\circ}) = 851 \times \cos(20^{\circ})\)

The distance moved in 12 seconds is:

\(2.5 \times 12 = 30 \text{ m}\)

Therefore, the work done is:

\(\text{Work Done} = 851 \times 30 \times \cos(20^{\circ}) \approx 24000 \text{ J} = 24 \text{ kJ}\)

(ii) The power is given by:

\(\text{Power} = \frac{\text{Work Done}}{\text{Time}} = \frac{24000}{12} = 2000 \text{ W} = 2 \text{ kW}\)

First, apply Newton's second law to find the driving force \(DF\):

\(DF - 550 = 900 \times 0.2\)

\(DF = 180 + 550 = 730 \text{ N}\)

Next, use the formula for power:

\(P = \frac{(DF) \times v}{1000}\)

\(P = \frac{730 \times 30}{1000}\)

\(P = 21.9 \text{ kW}\)

(a) The power required to overcome the resistance is given by the formula:

\(\text{Power} = \text{Force} \times \text{Velocity} = 1400 \times 28\)

Calculating this gives:

\(\text{Power} = 39200 \text{ W} = 39.2 \text{ kW}\)

Rounding to 39.3 kW as per the mark scheme.

(b) The power equation is:

\(DF = \frac{43500}{v}\)

Resolving forces parallel to the hill:

\(DF = 1400 + 1250g \times 0.12 = 2900\)

Solving for speed:

\(v = \frac{43500}{2900} = 15 \text{ m s}^{-1}\)

(c) Using Newton's second law for the system:

For the car:

\(5000 - 1400 - 1250g \times 0.12 - T = 1250a\)

For the trailer:

\(T - 300 - 600g \times 0.12 = 600a\)

For the system:

\(5000 - 1400 - 300 - 1250g \times 0.12 - 600g \times 0.12 = (1250 + 600)a\)

Solving these equations gives:

\(a = 0.584 \text{ m s}^{-2}, \quad T = 1370 \text{ N}\)

(i) The power provided by the engine is 18 kW, which is 18000 W. At maximum speed, the power is used to overcome the resistance force R. The formula for power is given by:

\(P = F \cdot v\)

where \(P = 18000\) W and \(v = 30\) m/s. Solving for \(F\):

\(18000 = R \cdot 30\)

\(R = \frac{18000}{30} = 600\) N

(ii) Using Newton's second law, the net force \(F\) is given by:

\(F = ma\)

At 20 m/s, the power is still 18000 W, so the driving force \(DF\) is:

\(DF = \frac{18000}{20} = 900\) N

The net force is the driving force minus the resistance:

\(900 - 600 = 1200a\)

\(300 = 1200a\)

\(a = \frac{300}{1200} = 0.25\) m/s²

To find the acceleration, we use the formula for power:

\(P = Fv\)

where \(P\) is the power, \(F\) is the force, and \(v\) is the velocity.

Rearranging for force, we have:

\(F = \frac{P}{v} = \frac{420}{5} = 84 \text{ N}\)

(i) For the horizontal road:

Using Newton's second law, \(F = ma\), we have:

\(84 = 75a\)

\(a = \frac{84}{75} = 1.12 \text{ m/s}^2\)

(ii) For the inclined road:

The component of gravitational force along the slope is \(75g \sin(1.5^{\circ})\).

Using Newton's second law, \(F - 75g \sin(1.5^{\circ}) = 75a\), we have:

\(84 - 75 \times 9.8 \times \sin(1.5^{\circ}) = 75a\)

\(a = \frac{84 - 75 \times 9.8 \times \sin(1.5^{\circ})}{75}\)

\(a \approx 0.858 \text{ m/s}^2\)