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

\(\text{KE gain} = \frac{1}{2} m v_B^2 = \frac{1}{2} \times 14000 \times (24)^2 = 4032 \times 10^3 \text{ J}\).

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

\(\text{PE loss} = m g \times AB \sin \theta = 14000 \times 9.8 \times 300 \sin \theta = 42 \times 10^6 \sin \theta \text{ J}\).

(ii) Using the work-energy principle, the work done by the driving force (WD) is equal to the gain in kinetic energy minus the loss in potential energy plus the work done against resistance:

\(5000 \times 10^3 = 4032 \times 10^3 - 42 \times 10^6 \sin \theta + 4800 \times 700\).

Simplifying gives:

\(5000 = 4032 - 42000 \sin \theta + 3360\).

Solving for \(\theta\):

\(5000 = 7392 - 42000 \sin \theta\).

\(42000 \sin \theta = 7392 - 5000\).

\(42000 \sin \theta = 2392\).

\(\sin \theta = \frac{2392}{42000}\).

\(\theta = \arcsin \left( \frac{2392}{42000} \right) \approx 3.3^{\circ}\).

(i) (a) The frictional force acting on B is given by \(F = \mu R = 0.7 \times 3\). The work done against the frictional force is \(\text{WD} = Fs = 2.1 \times 0.9 = 1.89 \text{ J}\).

(b) The loss of potential energy of the system is \(3 \times 0.9 = 2.7 \text{ J}\).

(c) The gain in kinetic energy of the system is given by \(\text{KE gain} = \text{loss in PE} - \text{WD by friction} = 2.7 - 1.89 = 0.81 \text{ J}\).

(ii) Using the kinetic energy equation, \(\frac{1}{2}(0.3 + 0.3)v_{\text{at break}}^2 = 0.81\). Solving for \(v_{\text{at break}}\), we find \(v_{\text{at break}}^2 = 2.7\).

Using the equation \(v_{\text{floor}}^2 = v_{\text{at break}}^2 + 2g \times 0.54\), we find \(v_{\text{floor}} = 3.67 \text{ m/s}\).

(i) The work done against resistance is calculated by considering the work done by the driving force and the gain in potential energy. The equation is:

\(4500 \times 1200 = 16000g \times 18 + \text{WD against resistance}\)

Solving for the work done against resistance:

\(\text{WD against resistance} = 4500 \times 1200 - 16000 \times 9.8 \times 18 = 2.52 \times 10^6 \text{ J}\)

(ii) The gain in kinetic energy from A to B is:

\(\frac{1}{2} \times 16000 \times (21^2 - 9^2) \text{ J} = 7680000 \text{ J}\)

The average force \(F\) is found using:

\(F = \frac{\text{KE gain} + 2000 \times 2400}{2400} = \frac{7680000 + 4800000}{2400} = 3200 \text{ N}\)

(iii) The power at A and B is calculated using:

\(P_A = (3200 + 1280) \times 9\)

\(P_B = (3200 - 1280) \times 21\)

Both powers are equal:

\(P_A = P_B = 40320 \text{ W}\)

(i) The potential energy (PE) loss when the ball falls 5 m is given by:

\(\text{PE loss} = 0.4g \times 5 = 20 \text{ J}\)

The initial kinetic energy (KE) when the ball starts moving upwards is:

\(\text{Initial KE}_{\text{up}} = 0.4g \times 5 - 12.8 = 7.2 \text{ J}\)

Using the energy conservation principle, the height \(h\) reached is given by:

\(0.4gh = 7.2\)

Solving for \(h\):

\(h = \frac{7.2}{0.4g} = 1.8 \text{ m}\)

(ii) The time taken to fall 5 m is calculated using:

\(5 = 0 + \frac{1}{2} g t_{\text{down}}^2\)

\(t_{\text{down}} = 1 \text{ s}\)

The time taken to reach the greatest height is:

\(0 = 6 - gt_{\text{up}}\) or \(1.8 = \frac{1}{2} g t_{\text{up}}^2\)

\(t_{\text{up}} = 0.6 \text{ s}\)

Total time is:

\(t_{\text{total}} = t_{\text{down}} + t_{\text{up}} = 1 + 0.6 = 1.6 \text{ s}\)

(i) The gain in kinetic energy (KE) of the system is the sum of the kinetic energies of both blocks:

\(\text{KE gain} = \frac{1}{2} \times 5 \times v^2 + \frac{1}{2} \times 16 \times v^2 = 10.5v^2 \text{ J}\)

(ii) (a) The loss of gravitational potential energy (PE) is given by the difference in potential energy of block B and block A:

\(\text{PE loss} = 16gx - 5g(x \sin 30^{\circ}) = 16gx - 5g \left( \frac{x}{2} \right) = 16gx - 2.5gx = 13.5gx\)

Substituting \(g = 10 \text{ m/s}^2\), we get:

\(\text{PE loss} = 135x \text{ J}\)

(ii) (b) The work done against the frictional force is calculated as:

\(R = 5g \cos 30^{\circ} = 5 \times 10 \times \frac{\sqrt{3}}{2} = 25 \sqrt{3} \text{ N}\)

Frictional force \(F = \frac{1}{\sqrt{3}} \times 25 \sqrt{3} = 25 \text{ N}\)

Work done against friction is:

\(\text{Work done} = F \times x = 25x \text{ J}\)

(iii) Using the energy principle:

\(\text{Gain in KE} = \text{Loss in PE} - \text{Work done against friction}\)

\(10.5v^2 = 135x - 25x\)

\(10.5v^2 = 110x\)

Multiplying both sides by 2:

\(21v^2 = 220x\)