(a) Use the equation of motion: \(s = ut + \frac{1}{2} a t^2\). Given \(u = 0\), \(s = 2\), and \(t = 5\), we have:

\(2 = \frac{1}{2} \times a \times 25\)

Solving for \(a\), we get \(a = 0.16 \text{ ms}^{-2}\).

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

\(R = 5g - X \sin 30\)

Using Newton's second law horizontally:

\(X \cos 30 - F = 5a\)

Where \(F = 0.4R\). Substituting \(R\) and \(F\):

\(X \cos 30 - 0.4(5g - X \sin 30) = 5 \times 0.16\)

Solving for \(X\), we find \(X = 19.5 \text{ N (3sf)}\).

(c) For limiting equilibrium, \(R = 5g - 25 \sin 30\). Calculating \(R\):

\(R = 37.5\)

The frictional force \(F\) is:

\(F = 25 \cos 30 = \frac{25 \sqrt{3}}{2}\)

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

\(\mu = \frac{F}{R} = 0.577 \text{ (3sf)}\)