(i) The work done by the force as the load moves from A to M is calculated using the formula:

\(\text{Work Done} = Fd \cos \theta\)

where \(F = 25 \text{ N}\), \(d = 40 \text{ m}\) (since M is the midpoint of AB), and \(\theta = 30^{\circ}\).

\(\text{Work Done} = 25 \times 40 \times \cos 30^{\circ} = 866 \text{ J}\)

(ii) For the motion from M to B, the work done by the force is given by:

\(50 \times 40 \cos 30^{\circ} = 866 + \text{KE gain}\)

Solving for the kinetic energy gain:

\(\text{KE gain} = 866 \text{ J}\)

The gain in kinetic energy is also given by:

\(\frac{1}{2} m (v^2 - u^2) = 866\)

where \(m = 35 \text{ kg}\), \(u = 1.2 \text{ m s}^{-1}\), and \(v\) is the final speed.

\(\frac{1}{2} \times 35 \times (v^2 - 1.2^2) = 866\)

\(17.5 \times (v^2 - 1.44) = 866\)

\(v^2 - 1.44 = \frac{866}{17.5}\)

\(v^2 = \frac{866}{17.5} + 1.44\)

\(v^2 = 49.4971\)

\(v = \sqrt{49.4971} \approx 7.14 \text{ m s}^{-1}\)