(i) The loss of kinetic energy (KE) is calculated using the formula:

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

where \(m = 20 \text{ kg}\), \(u = 2.5 \text{ m/s}\), and \(v = 1.5 \text{ m/s}\).

\(\text{KE loss} = \frac{1}{2} \times 20 \times (2.5^2 - 1.5^2) = 40 \text{ J}\)

The gain in potential energy (PE) is calculated using:

\(\text{PE gain} = mgL \sin \alpha\)

where \(g = 9.8 \text{ m/s}^2\), \(L = 10 \text{ m}\), and \(\alpha = 4.5^{\circ}\).

\(\text{PE gain} = 20 \times 9.8 \times 10 \times \sin 4.5^{\circ} = 157 \text{ J}\)

(ii) The work done by the pulling force is given by:

\(\text{Work done} = \text{PE gain} - \text{KE loss} + \text{work against resistance}\)

\(\text{Work done} = 157 - 40 + 50 = 167 \text{ J}\)

(iii) The magnitude of the pulling force \(F\) is found using:

\(\text{Work done} = F \cdot L \cdot \cos 15^{\circ}\)

\(167 = F \times 10 \times \cos 15^{\circ}\)

\(F = \frac{167}{10 \times \cos 15^{\circ}} = 17.3 \text{ N}\)

(i) (a) The potential energy (PE) loss is equal to the kinetic energy (KE) gain. Therefore,

\(0.2g(3 - h) = 1.6\)

Solving for \(h\), we get:

\(h = 2.2 \text{ m}\)

(b) The kinetic energy of the particle immediately before it reaches the ground is given by the total potential energy loss from \(A\) to the ground:

\(\text{KE} = 0.2g \times 3 = 6 \text{ J}\)

(c) The ratio \(v_G : v_B\) can be found using the relationship:

\(\left( \frac{v_G}{v_B} \right)^2 = \frac{3}{3 - 2.2}\)

Thus,

\(\frac{v_G}{v_B} = \sqrt{\frac{3}{0.8}} = 1.94\)

(ii) Given \(v_G : v_B = 2.55\), we use the relationship:

\(\frac{H}{H - 2.2} = 2.55^2\)

Solving for \(H\), we find:

\(H = 2.6 \text{ m}\)

(a) For limiting equilibrium, the tension in the string is equal to the weight of A:

\(T = 4g\)

The normal reaction force on B is given by:

\(R = 3g \cos 30^{\circ}\)

Resolving forces parallel to the plane for B gives:

\(F = T - 3g \sin 30^{\circ}\)

Using the frictional force equation \(F = \mu R\), we have:

\(4g - 3g \sin 30^{\circ} = \mu (3g \cos 30^{\circ})\)

Solving for \(\mu\):

\(\mu = \frac{4g - 3g \sin 30^{\circ}}{3g \cos 30^{\circ}} = 0.962\)

(b) Using conservation of energy, the potential energy lost by A is equal to the kinetic energy gained by both particles:

\(4gx - 3gx \sin 30^{\circ} = \frac{1}{2} \times 4 \times v^2 + \frac{1}{2} \times 3 \times v^2\)

Solving for \(v\):

\(4g - 3g \sin 30^{\circ} = \frac{7}{2} v^2\)

\(v^2 = \frac{100}{21}\)

\(v = \sqrt{\frac{100}{21}} \approx 2.18 \text{ m/s}\)

(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.

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

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

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

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

\(v^2 = 49.4971\)

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

(i) Using the conservation of energy, the kinetic energy (KE) loss is equal to the potential energy (PE) gain:

\(\frac{1}{2} m u^2 = m g (0.45)\)

Solving for \(u\):

\(\frac{1}{2} \times 0.3 \times u^2 = 0.3 \times 9.8 \times 0.45\)

\(u^2 = 2 \times 9.8 \times 0.45\)

\(u = 3 \text{ m/s}\)

(ii) The potential energy gain is calculated as:

\(\text{PE gain} = \frac{1}{2} \times 0.3 \times 3^2 - 0.39\)

\(\text{PE gain} = 0.96 \text{ J}\)

Using \(\text{PE} = mgh\):

\(0.3 \times 9.8 \times h = 0.96\)

\(h = \frac{0.96}{0.3 \times 9.8}\)

\(h = 0.32 \text{ m}\)