(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:

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

(a) Using the equation of motion \(s = ut + \frac{1}{2} a t^2\), where \(u = 0\), \(s = 1.44\) m, and \(a = 2\) m/s\(^2\):

\(1.44 = 0 + \frac{1}{2} \times 2 \times t^2\)

\(1.44 = t^2\)

\(t = \sqrt{1.44} = 1.2 \text{ s}\)

(b) Resolve forces vertically to find the normal reaction \(R\):

\(R = 0.4g - 3 \times \frac{3}{5} = 0.4g - 3 \sin 36.9^\circ \approx 2.2 \text{ N}\)

Resolve forces horizontally using Newton's second law:

\(3 \times \frac{4}{5} - F = 3 \cos 36.9^\circ - F = 0.4 \times 2\)

\(F = 1.6 \text{ N}\)

Using \(\mu = \frac{F}{R}\):

\(\mu = \frac{1.6}{2.2} \approx 0.727\)

The crate is moving at a constant speed, so the net force acting on it is zero. The pulling force is balanced by the frictional force.

The frictional force \(F\) is given by \(F = \mu R\), where \(R\) is the normal reaction force.

Since the crate is on horizontal ground, \(R = mg\), where \(m = 500 \text{ kg}\) and \(g = 9.8 \text{ m/s}^2\).

Thus, \(R = 500 \times 9.8 = 4900 \text{ N}\).

The pulling force is 2500 N, so \(2500 = \mu \times 4900\).

Solving for \(\mu\), we get \(\mu = \frac{2500}{4900} = 0.5\).

(i) Using the equation of motion \(s = ut + \frac{1}{2}at^2\), where \(u = 0\), \(s = 4.5\), and \(t = 5\), we have:

\(4.5 = 0 + \frac{1}{2} \times a \times 5^2\)

Solving for \(a\), we get \(a = 0.36 \text{ m/s}^2\).

Resolving horizontally, the net force is \(6 \times \frac{24}{25} - F = 3 \times 0.36\).

Solving for \(F\), we find \(F = 4.68 \text{ N}\).

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

\(R = 3g - 6 \sin 16.3 = 3g - 6 \times \frac{7}{25} = 28.32 \text{ N}\)

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

\(4.68 = \mu \times 28.32\)

Solving for \(\mu\), we get \(\mu = 0.165\).

(iii) The velocity at \(t = 5\) is \(v = 5 \times 0.36 = 1.8 \text{ m/s}\).

Using \(3a = -0.165 \times 3g\), we solve for \(t\) when \(v = 0\):

\(0 = 1.8 - 0.165g t\)

Solving for \(t\), we find \(t = 1.09 \text{ s}\).

The total time is \(5 + 1.09 = 6.09 \text{ s}\).

Let the mass of the block be denoted by \(m\).

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

\(R = mg + 50 \sin 20^{\circ}\)

The frictional force \(F\) is given by:

\(F = 0.3(mg + 50 \sin 20^{\circ})\)

Since the block moves at constant speed, the horizontal component of the applied force equals the frictional force:

\(50 \cos 20^{\circ} = 0.3(mg + 50 \sin 20^{\circ})\)

Solving for \(m\):

\(50 \cos 20^{\circ} = 0.3mg + 0.3 \times 50 \sin 20^{\circ}\)

\(50 \cos 20^{\circ} - 0.3 \times 50 \sin 20^{\circ} = 0.3mg\)

\(m = \frac{50 \cos 20^{\circ} - 0.3 \times 50 \sin 20^{\circ}}{0.3g}\)

Using \(g = 9.8 \text{ m/s}^2\), calculate \(m\):

\(m \approx 14.0 \text{ kg}\)