(i) Applying Newton's second law to particle B, we have:

\(T - 1.6g = 1.6a\)

For particle A:

\(0.4g - T = 0.4a\)

Adding these equations gives:

\(1.2g = 2a\)

Thus, the acceleration \(a = 6 \text{ m/s}^2\).

Substituting back to find \(T\):

\(T = 1.6g - 1.6a = 1.6 \times 10 - 1.6 \times 6 = 6.4 \text{ N}\)

The work done by tension is:

\(\text{Work done} = T \times 1.2 = 6.4 \times 1.2 = 7.68 \text{ J}\)

(ii) Using the work-energy principle for particle A after B reaches the floor:

\(\frac{1}{2} \times 1.6 \times v^2 = 1.6 \times g \times 1.2 - 7.68\)

\(v^2 = 14.4\)

Using the equation of motion for A:

\(v^2 = 2gh\)

\(14.4 = 2 \times 10 \times h\)

\(h = 0.72 \text{ m}\)

The greatest height above the floor is:

\(H = 2 \times 1.2 + h = 2.4 + 0.72 = 3.12 \text{ m}\)

To find the work done against resistance, we use the work-energy principle. The change in mechanical energy is equal to the work done against resistance.

The initial kinetic energy (KE) at the top is given by:

\(KE_{\text{top}} = \frac{1}{2} \times 15 \times 2^2 = 30 \text{ J}\)

The final kinetic energy (KE) at the bottom is:

\(KE_{\text{bottom}} = \frac{1}{2} \times 15 \times 4^2 = 120 \text{ J}\)

The change in potential energy (PE) as the block descends 1.6 m is:

\(\Delta PE = 15g \times 1.6 = 240 \text{ J}\)

According to the work-energy principle:

\(\Delta KE + \Delta PE = \text{Work done against resistance} + \text{Work done by gravity}\)

Substituting the values:

\(120 - 30 + 240 = 120 + W\)

Solving for \(W\):

\(240 + 30 = 120 + W\)

\(W = 150 \text{ J}\)

The gain in kinetic energy (KE) of the box is given by:

\(\text{Gain in KE} = \frac{1}{2} \times 25 \times 3^2 = 112.5 \text{ J}\)

The work done (WD) by the pulling force is:

\(\text{WD by pulling force} = 220 \times 15 \times \cos \alpha\)

According to the work-energy principle, the work done by the pulling force is equal to the gain in kinetic energy plus the work done against resistance:

\(220 \times 15 \times \cos \alpha = 112.5 + 3000\)

Solving for \(\cos \alpha\):

\(3300 \times \cos \alpha = 3112.5\)

\(\cos \alpha = \frac{3112.5}{3300}\)

\(\cos \alpha \approx 0.943\)

\(\alpha = \cos^{-1}(0.943) \approx 19.4^\circ\)

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

The gain in potential energy is given by:

\(\text{Gain in PE} = mgh = 15,000 \times 9.8 \times 16\)

\(= 2.352 \times 10^6 \text{ J}\)

The work done against resistance is:

\(\text{WD against resistance} = 1800 \times 1440\)

\(= 2.592 \times 10^6 \text{ J}\)

The total work done by the driving force is the sum of these two:

\(\text{Total work done} = 2.352 \times 10^6 + 2.592 \times 10^6\)

\(= 4.944 \times 10^6 \text{ J}\)

According to the mark scheme, the work done is approximately:

\(4.99 \times 10^6 \text{ J}\)

(ii) To find the distance from the top of the hill to point P, we use the work-energy principle.

The work done by the engine is given as 5030 kJ, which is:

\(5030 \times 10^3 \text{ J}\)

The increase in kinetic energy (KE) is:

\(\Delta KE = \frac{1}{2} m (v^2 - u^2) = \frac{1}{2} \times 15,000 \times (24^2 - 15^2)\)

\(= 2,925,000 \text{ J}\)

The work done against resistance is:

\(1600 \times d\)

Equating the work done by the engine to the increase in KE plus the work done against resistance:

\(5030 \times 10^3 = 2,925,000 + 1600d\)

Solving for \(d\):

\(5030 \times 10^3 - 2,925,000 = 1600d\)

\(2,105,000 = 1600d\)

\(d = \frac{2,105,000}{1600}\)

\(d = 1500 \text{ m}\)

(i) The work done by a force is given by \(W = Fd \cos \theta\), where \(F\) is the force, \(d\) is the distance moved, and \(\theta\) is the angle between the force and the direction of movement.

The work done by the 30 N force is \(30 \times 20 \times 0.6\).

The work done by the 40 N force is \(40 \times 20 \times 0.8\).

Total work done = \(30 \times 20 \times 0.6 + 40 \times 20 \times 0.8 = 1000 \text{ J}\).

(ii) Since the block is moving with constant speed, the net force in the x-direction is zero. The frictional force \(F_f = \mu W\), where \(\mu = \frac{5}{8}\) and \(W\) is the weight of the block.

The equation for equilibrium in the x-direction is:

\(30 \times 0.6 + 40 \times 0.8 - \frac{5}{8}W = 0\).

Solving for \(W\):

\(18 + 32 - \frac{5}{8}W = 0\)

\(50 = \frac{5}{8}W\)

\(W = \frac{50 \times 8}{5} = 80 \text{ N}\).