Resolve the forces perpendicular to the plane:

\(R = 12g \cos 25^{\circ} + P \sin 25^{\circ}\)

Resolve the forces parallel to the plane:

\(P \cos 25^{\circ} = F + 12g \sin 25^{\circ}\)

Since the particle is on the point of moving, the frictional force \(F\) is at its maximum value:

\(F = \mu R = 0.8R\)

Substitute \(F = 0.8R\) into the parallel force equation:

\(P \cos 25^{\circ} = 0.8R + 12g \sin 25^{\circ}\)

Substitute \(R = 12g \cos 25^{\circ} + P \sin 25^{\circ}\) into the equation:

\(P \cos 25^{\circ} = 0.8(12g \cos 25^{\circ} + P \sin 25^{\circ}) + 12g \sin 25^{\circ}\)

Simplify and solve for \(P\):

\(P \cos 25^{\circ} = 9.6g \cos 25^{\circ} + 0.8P \sin 25^{\circ} + 12g \sin 25^{\circ}\)

\(P \cos 25^{\circ} - 0.8P \sin 25^{\circ} = 9.6g \cos 25^{\circ} + 12g \sin 25^{\circ}\)

\(P(\cos 25^{\circ} - 0.8 \sin 25^{\circ}) = 9.6g \cos 25^{\circ} + 12g \sin 25^{\circ}\)

\(P = \frac{9.6g \cos 25^{\circ} + 12g \sin 25^{\circ}}{\cos 25^{\circ} - 0.8 \sin 25^{\circ}}\)

Substitute \(g = 9.8\):

\(P = \frac{9.6 \times 9.8 \times \cos 25^{\circ} + 12 \times 9.8 \times \sin 25^{\circ}}{\cos 25^{\circ} - 0.8 \times \sin 25^{\circ}}\)

\(P = 242\)

(a)(i) Resolve forces parallel to the plane:

\(T \cos 20^{\circ} - 5 - 2g \sin 30^{\circ} = 2 \times 1.2\)

Solving for T:

\(T \cos 20^{\circ} = 5 + 2g \sin 30^{\circ} + 2 \times 1.2\)

\(T = \frac{5 + 2g \sin 30^{\circ} + 2 \times 1.2}{\cos 20^{\circ}}\)

\(T = 18.5 \text{ N}\)

(a)(ii) Resolve forces perpendicular to the plane:

\(R = 2g \cos 30^{\circ} - T \sin 20^{\circ}\)

\(Using F = μR:\)

\(5 = \mu (2g \cos 30^{\circ} - T \sin 20^{\circ})\)

Solving for μ:

\(\mu = \frac{5}{2g \cos 30^{\circ} - T \sin 20^{\circ}}\)

\(\mu = 0.455\)

(b) Calculate the maximum friction force:

\(\text{Max } F = 0.8 \times (2g \cos 30^{\circ} - 15 \sin 20^{\circ})\)

\(= 9.7521 \ldots\)

Net force up the plane:

\(15 \cos 20^{\circ} - 2g \sin 30^{\circ} = 4.0953 \ldots\)

Since 4.0953 < 9.7521, the block does not move up the plane.

(i) Resolve forces down the plane:

\(20 + 5g \, \sin 10^\circ - F = 0\)

\(R = 5g \, \cos 10^\circ\)

\(\mu = \frac{F}{R} = \frac{20 + 8.6824}{49.24}\)

The coefficient of friction \(\mu\) is 0.582.

(ii) Using Newton's second law:

\(5g \, \sin 10^\circ - 0.582 \times 49.24 = 5a\)

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

\(0 = 2.5^2 - 2 \times 4s\)

The distance moved \(s\) is 0.781 m.

Alternative method for part (ii):

Potential energy loss:

\(5g \, d \, \sin 10^\circ\)

\(\frac{1}{2} \times 5 \times 2.5^2 + 5g \, d \, \sin 10^\circ = 0.582 \times 5g \, d \, \cos 10^\circ\)

The distance moved \(d\) is 0.781 m.

First, calculate the loss in kinetic energy (KE):

\(\text{KE loss} = \frac{1}{2} \times 12000 \times (24^2 - 16^2) = 3000000 \text{ J}\).

Next, calculate the gain in potential energy (PE):

\(\text{PE gain} = 12000 \times 9.8 \times 25 = 2940000 \text{ J}\).

The work done (WD) against resistance is:

\(\text{WD against resistance} = 7500 \times 500 = 3750000 \text{ J}\).

The work done by the driving force (DF) is given by:

\(\text{WD by DF} = \text{PE gain} - \text{KE loss} + \text{WD against resistance}\).

\(\text{WD by DF} = 2940000 - 3000000 + 3750000 = 3690000 \text{ J}\).

The driving force is then:

\(\text{Driving force} = \frac{3690000}{500} = 9660 \text{ N}\).

Using Newton's second law, the equation of motion for the box is:

\(P - 8g \sin 5^{\circ} - F = 8a\)

For \(P = 7X\) and acceleration \(a = 0.15 \text{ m/s}^2\):

\(7X - 8g \sin 5^{\circ} - F = 8 \times 0.15\)

For \(P = 8X\) and acceleration \(a = 1.15 \text{ m/s}^2\):

\(8X - 8g \sin 5^{\circ} - F = 8 \times 1.15\)

Solving these equations simultaneously gives:

\(X = 8\)

Substitute \(X = 8\) into one of the equations to find \(F\):

\(F = 56 - 8g \sin 5^{\circ} - 8 \times 0.15\)

\(F = 47.8 \text{ N}\)

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

\(R = 8g \cos 5^{\circ}\)

\(R = 79.695 \text{ N}\)

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

\(\mu = \frac{F}{R} = \frac{47.8}{79.7}\)

\(\mu = 0.600\)