(i) The acceleration of the particle down the slope is given by the component of gravitational acceleration along the slope:

\(a = g \sin 30^{\circ} = 5 \text{ m/s}^2\).

Using the equation of motion \(v = u + at\), where \(u = 0\), \(v = 2.5 \text{ m/s}\), and \(a = 5 \text{ m/s}^2\), we have:

\(2.5 = 0 + 5t\)

Solving for \(t\), we get \(t = 0.5 \text{ s}\).

(ii) When the particle reaches the horizontal ground, it has a velocity \(v^2 = 0 + 2 \times 5 \times 3 = 30 \text{ m}^2/\text{s}^2\).

On the horizontal ground, the frictional force provides a deceleration \(a\) given by \(1 = 0.5a\), so \(a = -2 \text{ m/s}^2\).

Using \(v^2 = u^2 + 2as\) with \(v = 0\), \(u^2 = 30\), and \(a = -2\), we have:

\(0 = 30 + 2(-2)s\)

Solving for \(s\), we get \(s = 7.5 \text{ m}\).

First, calculate the frictional force \(F\) using the formula:

\(F = 0.25 \left( 6.1 \times \frac{60}{61} \right) = 1.5 \text{ N}\)

Using Newton's second law, the equation of motion along the slope is:

\(6.1 \times \frac{11}{61} - 0.25 \times 6.1 \times \frac{60}{61} = 0.61a\)

Solving for \(a\):

\(a = \frac{-40}{61} = -0.656 \text{ m/s}^2\)

Using the equation \(0 = v_A^2 + 2as\) to find the distance \(s\):

\(0 = 2^2 + 2 \times \left(-0.656\right) \times s\)

\(s = \frac{4}{2 \times 0.656} = 3.05 \text{ m}\)

Thus, the block moves 3.05 m from A before it comes to rest.

(i) (a) The acceleration of P from A to B is given by \(g \sin \alpha\). Substituting \(g = 9.8\) m s^-2 and \(\sin \alpha = 0.28\), we have:

\(a = 9.8 \times 0.28 = 2.8 \text{ m s}^{-2}\)

(b) For the motion from B to C, using Newton's second law:

\(mg \times 0.28 - 0.5mg \times 0.96 = ma\)

Solving for \(a\):

\(a = \frac{mg(0.28 - 0.5 \times 0.96)}{m} = -2 \text{ m s}^{-2}\)

(ii) Using \(v^2 = u^2 + 2as\) for AB and BC and \(AB + BC = 5\):

\(v_B^2 = 2 \times 2.8(AB)\)

\(2^2 = 5.6(AB) - 2 \times 2(5 - AB)\)

Solving gives \(AB = 2.5 \text{ m}\).

(iii) Using \(t = \frac{2s}{u+v}\) for AB and BC:

\(T = 2 \times 2.5 \div (0 + \sqrt{14}) + 2 \times 2.5 \div (\sqrt{14} + 2)\)

The time taken is 2.21 s.

Using Newton's second law, the acceleration down the slope is given by:

\(a = g (\sin 14^{\circ} - 0.02 \cos 14^{\circ})\)

Substituting the values, we get:

\(a = 9.8 (\sin 14^{\circ} - 0.02 \cos 14^{\circ}) \approx 2.225 \text{ m s}^{-2}\)

Using the equation of motion:

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

where \(u = 8 \text{ m s}^{-1}\), \(a = 2.225 \text{ m s}^{-2}\), and \(s = 50 \text{ m}\).

Substitute the values:

\(v^2 = 8^2 + 2 \times 2.225 \times 50\)

\(v^2 = 64 + 222.5\)

\(v^2 = 286.5\)

\(v = \sqrt{286.5} \approx 16.9 \text{ m s}^{-1}\)

Thus, the speed of the sledge and the man when they reach the bottom of the track is 16.9 m s^-1.

(a) Resolve forces parallel and perpendicular to the plane. The normal reaction \(R = 5g \cos 30\). The force equation parallel to the plane is:

\(40 - 5g \sin 30 - F > 0\)

where \(F = \mu R = \mu \times 5g \cos 30\).

Substitute \(F\) into the inequality:

\(40 - 5g \sin 30 - \mu \times 5g \cos 30 > 0\)

Simplify to find \(\mu\):

\(\mu < \frac{1}{5} \sqrt{3}\)

(b) Resolve forces horizontally and vertically. The normal reaction \(R = 5g \cos 30 + 40 \sin 30\). The frictional force \(F = 40 \cos 30 - 5g \sin 30\).

Substitute \(F = \mu R\) and solve for \(\mu\):

\(\mu = \frac{40 \cos 30 - 5g \sin 30}{5g \cos 30 + 40 \sin 30}\)

Calculate \(\mu\) to 3 decimal places:

\(\mu = 0.152\)