(i) Using Newton's second law along the plane, the equation is:

\(-F - 0.6g\sin 18^{\circ} = 0.6(-4)\)

Solving for the frictional force \(F\):

\(F = 0.6g\sin 18^{\circ} - 0.6 \times 4\)

\(F = 0.546\, \text{N}\)

For the normal component, resolve forces perpendicular to the plane:

\(R = 0.6g\cos 18^{\circ}\)

\(R = 5.71\, \text{N}\)

The coefficient of friction \(\mu\) is given by:

\(\mu = \frac{F}{R} = \frac{0.546}{5.71} = 0.096\)

(ii) When the particle moves down the plane, using Newton's second law:

\(0.6g\sin 18^{\circ} - 0.546 = 0.6a\)

Solving for \(a\):

\(a = \frac{0.6g\sin 18^{\circ} - 0.546}{0.6}\)

\(a = 2.18\, \text{m s}^{-2}\)

(i) The normal force R is given by:

\(R = mg \cos 21^{\circ} = 9.336m\)

Using Newton's second law for motion up the plane:

\(a = -5\)

The equation of motion is:

\(-mg \sin 21^{\circ} - F = -5m\)

Solving for F gives:

\(F = 1.416m\)

(ii) The coefficient of friction \(\mu\) is:

\(\mu = \frac{F}{R} = \frac{1.416m}{9.336m} = 0.152\)

(iii) Using the equation of motion for the particle moving down the plane:

\(m g \sin 21^{\circ} - 1.416m = ma\)

Solving for acceleration \(a\):

\(a = 2.167 \text{ m/s}^2\)

Using the equation \(v^2 = 2as\) with \(s = 10\):

\(v^2 = 2(2.167)(10)\)

Solving for \(v\):

\(v = 6.58 \text{ m/s}\)

(a) In limiting equilibrium, the frictional force \(F\) is equal to \(\mu R\), where \(R\) is the normal reaction force. The forces acting on the particle are its weight \(0.4g\) and the normal reaction \(R\). The component of the weight perpendicular to the plane is \(0.4g \cos 30^{\circ}\), and the component parallel to the plane is \(0.4g \sin 30^{\circ}\).

Since the particle is in limiting equilibrium, \(F = \mu R = 0.4g \sin 30^{\circ}\).

The normal reaction \(R = 0.4g \cos 30^{\circ}\).

Thus, \(\mu = \frac{0.4g \sin 30^{\circ}}{0.4g \cos 30^{\circ}} = \frac{\sin 30^{\circ}}{\cos 30^{\circ}} = \tan 30^{\circ} = \frac{1}{3} \sqrt{3}\).

(b) Applying Newton's second law along the plane: \(7.2 - 0.4g \sin 30^{\circ} - F = 0.4a\).

Substituting \(F = \mu R = \frac{1}{3} \sqrt{3} \times 0.4g \cos 30^{\circ}\), we find \(a = 8 \text{ m/s}^2\).

Using the equation of motion \(s = ut + \frac{1}{2}at^2\), where \(s = 1 \text{ m}\), \(u = 0\), and \(a = 8 \text{ m/s}^2\), we solve for \(t\):

\(1 = 0 + \frac{1}{2} \times 8 \times t^2\)

\(t^2 = \frac{1}{4}\)

\(t = 0.5 \text{ s}\).

First, use the equation of motion to find the acceleration:

\(s = \frac{1}{2} a t^2\)

\(2.25 = \frac{1}{2} a (1.5^2)\)

\(a = 2 \text{ m/s}^2\)

Next, calculate the normal reaction force \(R\):

\(R = mg \cos 30^{\circ}\)

Apply Newton's second law along the plane:

\(mg \sin 30^{\circ} - \mu mg \cos 30^{\circ} = ma\)

\(mg \sin 30^{\circ} - \mu mg \cos 30^{\circ} = 2m\)

Solving for \(\mu\):

\(\mu = \frac{mg \sin 30^{\circ} - 2m}{mg \cos 30^{\circ}}\)

\(\mu = \frac{g \sin 30^{\circ} - 2}{g \cos 30^{\circ}}\)

Substitute \(g = 9.8 \text{ m/s}^2\):

\(\mu = \frac{9.8 \times 0.5 - 2}{9.8 \times \frac{\sqrt{3}}{2}}\)

\(\mu = 0.346\)

(i) The potential energy gain is given by:

\(PE = mg(2.5 \sin 60^{\circ})\)

Using kinetic energy: \(KE = \frac{1}{2} mv^2\)

By conservation of energy:

\(\frac{1}{2} m (8^2 - v^2) = mg(2.5 \sin 60^{\circ})\)

Solving for \(v\), we find \(v = 4.55 \text{ m s}^{-1}\).

(ii) Using \(\frac{1}{2} mu^2 > mgh_{\text{max}}\),

\(\frac{1}{2} \times 8^2 > 10h_{\text{max}}\)

This shows \(h_{\text{max}} < 3.2 \text{ m}\).

(iii) Since \(BC\) is smooth and \(B\) and \(C\) are at the same height, energy is conserved, so speed at \(C\) is the same as at \(B\).

(iv) Work done against friction is \(1.4 \times 5.2\).

Using work-energy principle:

\(1.4 \times 5.2 = \frac{1}{2} \times 0.4 (v^2 - 4.55^2)\)

Solving for \(v\), we find \(v = 5.25 \text{ m s}^{-1}\).