(i) To find the speed of the ring as it reaches B, use the equation of motion:

\(v^2 = u^2 + 2as\)

where \(u = 0\) (initial speed), \(a = 2.5\) m/s² (acceleration), and \(s = 5\) m (distance).

\(v^2 = 0 + 2 \times 2.5 \times 5\)

\(v^2 = 25\)

\(v = 5\) m/s

(ii)(a) To find the speed of the ring at C, use energy conservation:

\(Potential energy loss = Kinetic energy gain\)

\(0.2 \times 10 \times 6 \times \sin 30^\circ = 0.5 \times 0.2 \times (v^2 - 5^2)\)

\(6 = 0.1(v^2 - 25)\)

\(6 = 0.1v^2 - 2.5\)

\(8.5 = 0.1v^2\)

\(v^2 = 85\)

\(v = 9.22\) m/s

(ii)(b) The greatest speed of the ring occurs at the lowest point of the semicircle:

\(Potential energy loss = Kinetic energy gain\)

\(0.2 \times 10 \times 6 = 0.5 \times 0.2 \times (v^2 - 5^2)\)

\(12 = 0.1(v^2 - 25)\)

\(12 = 0.1v^2 - 2.5\)

\(14.5 = 0.1v^2\)

\(v^2 = 145\)

\(v = 12.0\) m/s

(i) The change in kinetic energy (KE) is given by:

\(\Delta KE = \frac{1}{2} \times 800 \times (14^2 - 8^2) = 52800 \text{ J}\)

The change in gravitational potential energy (PE) is given by:

\(\Delta PE = 800 \times 10 \times 120 \times 0.15 = 144000 \text{ J}\)

(ii) The work done by the engine is:

\(\text{WD by engine} = 32000 \times 12 = 384000 \text{ J}\)

Using the work-energy principle:

\(384000 = 144000 + 52800 + \text{WD against } F\)

Solving for the work done against resistive forces:

\(\text{WD against } F = 384000 - 144000 - 52800 = 187200 \text{ J}\)

Rounding to 3 significant figures, the work done against resistive forces is 187000 J.

(i) The initial kinetic energy (KE) of the particle is given by the formula:

\(KE = \frac{1}{2} m v^2\)

Substituting the given values:

\(KE = \frac{1}{2} \times 0.4 \times 12^2 = 28.8 \text{ J}\)

(ii) To find the distance the particle moves up the plane, we use the conservation of energy principle. The loss in kinetic energy is equal to the gain in potential energy (PE).

The gain in potential energy is given by:

\(PE = mgh\)

where \(h\) is the height gained. Since the plane is inclined at 30°, the height \(h\) can be expressed in terms of the distance \(d\) traveled up the plane:

\(h = d \sin 30^{\circ}\)

Thus, the potential energy gain is:

\(PE = 0.4g(d \sin 30^{\circ})\)

Equating the kinetic energy loss to the potential energy gain:

\(28.8 = 0.4g(d \sin 30^{\circ})\)

\(28.8 = 0.4 \times 9.8 \times d \times 0.5\)

\(28.8 = 1.96d\)

Solving for \(d\):

\(d = \frac{28.8}{1.96} = 14.4 \text{ m}\)

(i) The forces acting on the box are the gravitational component down the slope \(F = 50g \sin 10^{\circ}\) and the normal reaction \(R = 50g \cos 10^{\circ}\). For the box to remain at rest, the frictional force must be at least equal to the component of gravity down the slope. Therefore, the coefficient of friction \(\mu\) must satisfy:

\(\mu \geq \frac{F}{R} = \frac{50g \sin 10^{\circ}}{50g \cos 10^{\circ}} = \tan 10^{\circ}\)

Calculating \(\tan 10^{\circ}\), we find \(\mu \geq 0.176\).

(ii) The work done by the girl is \(50 \times 5\). The potential energy loss is \(50g \times 10 \sin 10^{\circ}\). The work done against friction over 10 m is \(0.19 \times 50g \cos 10^{\circ} \times 10\). Using the energy method:

\(50 \times 5 + 50g \times 10 \sin 10^{\circ} - 0.19 \times 50g \cos 10^{\circ} \times 10 = 0.5 \times 50v^2\)

Solving for \(v\), we find \(v = 2.70 \text{ m/s}\).

(iii) On the 20° incline, the net force is \(50g \sin 20^{\circ} - 0.19 \times 50g \cos 20^{\circ}\). Using Newton's second law:

\(50g \sin 20^{\circ} - 0.19 \times 50g \cos 20^{\circ} = 50a\)

Solving for \(a\), we find \(a = 1.63 \text{ m/s}^2\).

(i) The work done by the pulling force is calculated using the formula:

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

Substituting the given values:

\(\text{Work Done} = 50 \times 20 \times \cos 10^{\circ} = 984.8 \text{ J}\)

(ii) Using the energy method, the work done by the driving force is equal to the kinetic energy gain plus the work done against resistance:

\(984.8 = \frac{1}{2} \times 25 \times v^2 + 30 \times 20\)

Solving for \(v\):

\(984.8 = 12.5v^2 + 600\)

\(384.8 = 12.5v^2\)

\(v^2 = 30.784\)

\(v = 5.55 \text{ ms}^{-1}\)

(iii) The greatest power is exerted when the speed is maximum:

\(\text{Power} = Fv\)

\(\text{Max Power} = 50 \cos 10^{\circ} \times 5.55 = 273 \text{ W}\)

(iv) Using Newton's second law up the plane:

\(50 \cos 10^{\circ} - 30 - 25g \sin 5^{\circ} = 25a\)

Solving for \(a\):

\(a = -0.102 \text{ ms}^{-2}\)

Using \(v = u + at\):

\(0 = 5.55 - 0.102t\)

\(t = \frac{5.55}{0.102} = 54.4 \text{ s}\)