(i) Using the formula for power, \(P = Fv\), where \(P = 36000\) W and \(F = 800\) N, we have:

\(36000 = 800v\)

\(v = \frac{36000}{800} = 45 \text{ m/s}\)

The distance \(AB\) is given by \(v \times t = 45 \times 120 = 5400 \text{ m}\).

(ii) Using Newton's second law, \(-800 = 900a\), we find \(a = -\frac{8}{9} \text{ m/s}^2\).

Using the equation \(v^2 = u^2 + 2as\), where \(u = 45 \text{ m/s}\), \(s = 450 \text{ m}\):

\(v^2 = 45^2 - \frac{16}{9} \times 450\)

\(v = 35 \text{ m/s}\)

(iii) The distance \(CD = 6637.5 - 5400 - 450 = 787.5 \text{ m}\).

Using \(v^2 = u^2 + 2as\), where \(u = 35 \text{ m/s}\), \(v = 0\):

\(0 = 35^2 - 2d \times 787.5\)

\(d = \frac{7}{9} = 0.778 \text{ m/s}^2\)

Using \(F = ma\), \(R = 900 \times \frac{7}{9} = 700 \text{ N}\).

(i) The driving force is given by the power divided by the speed: \(\frac{160}{5} = 32 \text{ N}\). Using Newton's Second Law, \(32 - 20 = m \times 0.15\). Solving for \(m\), we get \(m = \frac{12}{0.15} = 80 \text{ kg}\).

(ii) For the hill, resolve forces: \(\frac{300}{v} - 20 - 80g \sin 2^\circ = 0\). Solving for \(v\), we find \(v = 6.26 \text{ m s}^{-1}\).

(iii) The driving force at 90% of the speed is \(\frac{300}{0.9 \times 6.26} = 53.2 \text{ N}\). Using Newton's Second Law: \(53.2 - 20 - 80g \sin 2^\circ = 80a\). Solving for \(a\), we get \(a = 0.0666 \text{ m s}^{-2}\).

(i) The potential energy (PE) gain is calculated as:

\(\text{PE gain} = 50g \times 3.5 = 1750 \text{ J}\)

The work done (WD) by the crane is the sum of the PE gain and the work done against resistance:

\(\text{WD} = 50g \times 3.5 + 25 \times 3.5 = 1837.5 \text{ J}\)

Thus, the work done by the crane is 1837.5 J or 1840 J.

(ii) The power (P) of the crane is calculated using the formula:

\(P = \frac{\text{WD}}{t} = \frac{1837.5}{2} = 918.75 \text{ W}\)

Rounding gives the power as 919 W.

(a) Since the locomotive and truck are moving at a constant speed, the net force is zero. Therefore, the tension in the coupling is 0 N.

(b) The power produced by the locomotive's engine is given by the formula \(P = Fv\), where \(F\) is the force and \(v\) is the velocity. The resistance force is 0.2 N, and the velocity is 2 m/s. Thus, \(P = 0.2 \times 2 = 0.4 \text{ W}\).

(c) When the power is changed to 1.2 W, the driving force \(DF\) is \(\frac{1.2}{2} = 0.6 \text{ N}\). Using Newton's second law for the locomotive: \(DF - 0.2 - T = 0.8a\). For the truck: \(T = 0.4a\). For the system: \(DF - 0.2 = 1.2a\). Solving these equations gives \(T = \frac{2}{15} \text{ N}\).

(i) The power of the van's engine is given by the formula:

\(P = Fv\)

where \(F\) is the total resistance force and \(v\) is the velocity. The total resistance force is \(300 + 100 = 400\) N. Therefore,

\(P = 400 \times 25 = 10,000\) W.

(ii) The tension in the cable is equal to the resistance force on the trailer, which is 100 N.

(iii) When the van is moving up the hill, the new driving force is calculated as:

\(\text{Driving force} = \frac{25,000}{20} = 1,250\) N.

Using Newton's second law for the system of van and trailer:

\(1,250 - 300 - T - 3000 \sin 4^{\circ} = 3000a\)

or for the trailer:

\(T - 100 - 500 \sin 4^{\circ} = 500a\)

Solving these equations, we find:

\(T = 221\) N.