(i) The potential energy gain from A to B is given by:

\(\text{PE gain} = mgh = 1200 \times 9.8 \times 45\)

The work done against resistance is 360 kJ, so the total work done by the engine is:

\(\text{Work done} = \text{PE gain} + 360,000 = 900,000 \text{ J or 900 kJ}\)

(ii) The work done against resistance from B to C is:

\(\text{WD against resistance} = 360 \times \frac{\sin 5^\circ}{\sin 1^\circ} \text{ kJ} = 1798 \text{ kJ}\)

The kinetic energy gain is given by:

\(1660 + 540 - 1798 = \frac{1}{2} \times 1200 \times (v^2 - 15^2)\)

Solving for \(v\):

\(402,000 = \frac{1}{2} \times 1200 \times (v^2 - 225)\)

\(v = 29.9 \text{ m s}^{-1}\)

(iii) The ratio of the power developed by the car’s engine immediately after leaving B to the power developed immediately before reaching C is:

\(\frac{P_B}{P_C} = \left( \frac{DF_B}{DF_C} \right) \times \frac{v_B}{v_C} = 1.5 \times \frac{15}{29.9}\)

\(\text{Ratio} = 0.75\)

(i) To find the work done by the lorry's driving force, we calculate the gain in potential energy (PE) and the work done against resistance.

Gain in PE = mass × gravity × height = 15,000 kg × 9.8 m/s² × 500 m × \(\sin 2.5^{\circ}\)

\(Work done against resistance = resistance × distance = 800 N × 500 m\)

\(Total work done = Gain in PE + Work done against resistance = 3,271,454 J + 400,000 J = 3,670,000 J or 3670 kJ\)

(ii) For the return journey, we use the work-energy principle.

\(Work done by driving force = 2000 N × 500 m = 1,000,000 J\)

Gain in kinetic energy (KE) = \(\frac{1}{2} \times 15,000 \times (v^2 - 20^2)\)

\(Using the equation: Gain in KE = Loss in PE - Work done against resistance + Work done by driving force\)

\(\frac{1}{2} \times 15,000 \times (v^2 - 20^2) = 3,271,454 - 400,000 + 1,000,000\)

Solving for \(v\), we find \(v = 30.3 \text{ m/s}\)

We start by using the principle of conservation of energy. The loss in potential energy (PE) is equal to the gain in kinetic energy (KE) plus the work done (WD) against resistance.

The change in kinetic energy is given by:

\(\text{Change in KE} = \frac{1}{2} m (v^2 - u^2)\)

where \(m = 8 \text{ kg}\), \(v = 8 \text{ m/s}\), and \(u = 3 \text{ m/s}\).

\(\text{Change in KE} = \frac{1}{2} \times 8 \times (8^2 - 3^2) = \frac{1}{2} \times 8 \times (64 - 9) = \frac{1}{2} \times 8 \times 55 = 220 \text{ J}\)

The work done against resistance is 120 J, so the total energy change is:

\(\text{Total energy change} = 220 + 120 = 340 \text{ J}\)

The loss in potential energy is given by:

\(\text{PE loss} = mgh\)

where \(g = 9.8 \text{ m/s}^2\) and \(h\) is the height.

\(340 = 8 \times 9.8 \times h\)

\(h = \frac{340}{8 \times 9.8} = \frac{340}{78.4} \approx 4.25 \text{ m}\)

Therefore, the height of the top of the plane above the level of the bottom is 4.25 m.

(i) Using Newton's second law for A and B:

For A: \(T - 1.2g = 1.2a\)

For B: \(2g - T = 2a\)

Solving these equations simultaneously:

\(T - 1.2 \times 9.8 = 1.2a\)

\(19.6 - T = 2a\)

Add the equations: \(19.6 - 1.2 \times 9.8 = 3.2a\)

\(a = 2.5 \text{ m/s}^2\)

Substitute \(a\) back to find \(T\):

\(T = 1.2 \times 9.8 + 1.2 \times 2.5 = 15 \text{ N}\)

(ii) (a) Gain in potential energy:

\(\Delta PE = mgh = 1.2 \times 9.8 \times 1.5 = 18 \text{ J}\)

(b) Work done by tension:

\(W = T \times d = 15 \times 1.5 = 22.5 \text{ J}\)

(c) Gain in kinetic energy:

\(\Delta KE = W - \Delta PE = 22.5 - 18 = 4.5 \text{ J}\)

(iii) Time for string to become taut again:

Velocity of A when B hits the floor:

\(v = u + at = 0 + 2.5 \times 1.6 = 4 \text{ m/s}\)

Time for A to stop and return:

\(v = u - gt\)

\(0 = 4 - 9.8t\)

\(t = \frac{4}{9.8} \approx 0.4 \text{ s}\)

Total time: \(0.4 \times 2 = 0.8 \text{ s}\)

(i) Using conservation of energy, the potential energy at A is converted to kinetic energy at B. The potential energy at A is given by:

\(mgh = 0.5 \times 10 \times 1.8\)

The kinetic energy at B is:

\(\frac{1}{2} mv^2\)

Equating the potential energy to kinetic energy:

\(\frac{1}{2} v^2 = 10 \times 1.8\)

\(v^2 = 36\)

\(v = 6 \text{ m s}^{-1}\)

(ii) The work done against resistance is the loss of kinetic energy as the particle moves from B to C. The kinetic energy at B is:

\(\frac{1}{2} \times 0.5 \times 6^2\)

The kinetic energy at C is:

\(\frac{1}{2} \times 0.5 \times 5^2\)

The work done is the difference:

\(\text{WD} = \frac{1}{2} \times 0.5 \times (6^2 - 5^2)\)

\(\text{WD} = 2.75 \text{ J}\)