(i) Using Newton's second law, the frictional force is given by:

\(F = \mu R = \mu mg\)

where \(\mu\) is the coefficient of friction, \(R\) is the normal reaction, and \(m\) is the mass of the particle. The deceleration \(a\) is given by:

\(\mu mg = ma\)

Thus, \(a = \mu g\).

For particle P, \(\mu = 0.2\):

\(a_P = 0.2 \times 9.8 = 2 \text{ m s}^{-2}\)

For particle Q, \(\mu = 0.25\):

\(a_Q = 0.25 \times 9.8 = 2.5 \text{ m s}^{-2}\)

(ii) Using the equation of motion \(s = ut + \frac{1}{2}at^2\), where \(s_P = s_Q + 5\):

For P:

\(s_P = 8t - \frac{1}{2} \times 2 \times t^2 = 8t - t^2\)

For Q:

\(s_Q = 3t - \frac{1}{2} \times 2.5 \times t^2 = 3t - 1.25t^2\)

Equating and solving:

\(8t - t^2 = 3t - 1.25t^2 + 5\)

\(5t = 0.25t^2 + 5\)

\(t^2 - 5t + 5 = 0\)

Solving for \(t\):

\(t = \sqrt{120} - 10 \approx 0.95445\)

Using \(v = u + at\) for both P and Q:

For P:

\(v_P = 8 - 2 \times 0.95445 = 6.09 \text{ m s}^{-1}\)

For Q:

\(v_Q = 3 - 2.5 \times 0.95445 = 0.614 \text{ m s}^{-1}\)

(i) Resolve the forces in the x-direction:

\(100 \cos 30^{\circ} + 120 \cos 60^{\circ} - F \cos \alpha = 136\)

\(F \cos \alpha = 10.6025\)

Resolve the forces in the y-direction:

\(100 \sin 30^{\circ} - 120 \sin 60^{\circ} + F \sin \alpha = 0\)

\(F \sin \alpha = 53.9230\)

Using the equations:

\(F^2 = (F \cos \alpha)^2 + (F \sin \alpha)^2\)

Calculate \(F\) and \(\alpha\):

\(F = 55.0\) N, \(\alpha = 78.9^{\circ}\)

(ii) The magnitude of the frictional force is equal to the resultant force, which is 136 N. The normal reaction \(R = 40g\), where \(g\) is the acceleration due to gravity.

The coefficient of friction \(\mu = \frac{136}{40g} = 0.34\).

(i) In limiting equilibrium, the frictional force is equal to the tension in the string. Therefore, we have:

\(T = \mu R\)

Given that the tension \(T = 24 \text{ N}\) and the weight \(W = 30 \text{ N}\), the normal reaction \(R = W = 30 \text{ N}\). Thus:

\(24 = \mu \times 30\)

Solving for \(\mu\):

\(\mu = \frac{24}{30} = 0.8\)

(ii) When the block is in motion, resolve the forces vertically and horizontally. The vertical component of the tension is \(25 \sin 30^\circ\) and the horizontal component is \(25 \cos 30^\circ\).

The normal reaction \(R\) is given by:

\(R = 30 - 25 \sin 30^\circ\)

The frictional force \(F\) is:

\(F = \mu R = 0.8 \times (30 - 25 \sin 30^\circ) = 14 \text{ N}\)

Using Newton's second law horizontally:

\(25 \cos 30^\circ - F = \frac{30}{g} a\)

Substitute \(F = 14 \text{ N}\) and solve for \(a\):

\(25 \cos 30^\circ - 14 = \frac{30}{g} a\)

\(a = 2.55 \text{ m/s}^2\)

(i) The frictional force \(F\) required to move the block horizontally is given by \(F = \mu W\), where \(\mu = 1.25\) and \(W\) is the weight of the block. Therefore, \(F = 1.25W\). Since the vertical force required to lift the block is \(W\), and \(F = 1.25W\), the vertical force \(W\) is less than the horizontal force \(F\).

(ii) Using Newton's second law, the net force \(F_{\text{net}} = ma\), where \(m\) is the mass and \(a\) is the acceleration. The mass \(m = \frac{W}{g} = \frac{60}{10} = 6\) kg (assuming \(g = 10\) m/s\(^2\)).

The net force is also given by \(P - F = ma\), where \(F = 1.25 \times 60 = 75\) N is the frictional force.

Thus, \(P - 75 = 6 \times 4\).

Solving for \(P\), we get \(P = 24 + 75 = 99\) N.

(i) Resolve forces in the x-direction:

\(F_x = 6.1 - 5 \times 0.28 = 0\)

Resolve forces in the y-direction:

\(F_y = 4.8 - 5 \times 0.96 = 0\)

The frictional force acts parallel to the x-axis and to the right. Therefore, \(F_y = 0 \Rightarrow F = F_x\). The frictional force has magnitude 7.5 N.

(ii) Using \(F = \mu R\) and \(R = mg\), we have:

\(\mu = \frac{7.5}{1.25 \times 10}\)

The coefficient of friction is 0.6.

(iii) Applying Newton's second law:

\([7.5 - 8.6 - 1.4 = 1.25a \Rightarrow a = -2]\)

The magnitude of acceleration is 2 m/s² and the direction is parallel to the x-axis and to the left.