(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:

\(21v^2 = 220x\)