(i) To find the speed of the sledge at the bottom of the slope, we use the conservation of energy. The potential energy (PE) lost is converted into kinetic energy (KE). The loss in potential energy is given by:

\(\text{PE loss} = mg \times 100 \sin 20^{\circ}\)

Using the equation for kinetic energy gain:

\(\frac{1}{2}mv^2 - \frac{1}{2}m \times 5^2 = mg \times 100 \sin 20^{\circ}\)

Solving for \(v\), we find:

\(v = 26.6 \text{ m/s}\)

(ii) When resistance is present, the work done against resistance is 8500 J. The kinetic energy at the bottom is given by:

\(\text{KE} = \frac{1}{2}m \times 21^2 - \frac{1}{2}m \times 5^2\)

\(\text{KE} = \pm (0.5m \times 441 - 0.5m \times 25) = \pm 208m\)

The equation for energy balance is:

\(mg \times 100 \sin 20^{\circ} = 8500 + 208m\)

Solving for \(m\), we find:

\(m = 63.4 \text{ kg}\)

The total mass of the cyclist and bicycle is 105 kg.

The initial speed is 5 m/s and the final speed is 10 m/s.

The kinetic energy gain is calculated as:

\(\text{KE gain} = \frac{1}{2} \times 105 \times (10^2 - 5^2) = \frac{1}{2} \times 105 \times (100 - 25) = \frac{1}{2} \times 105 \times 75 = 3937.5 \text{ J}\)

The work done against resistance is:

\(\text{WD against Resistance} = 50 \times 40 = 2000 \text{ J}\)

The total work done is the sum of the kinetic energy gain and the work done against resistance:

\(\text{Total WD} = 3937.5 + 2000 = 5937.5 \text{ J}\)

(i) To find the distance travelled, use the equation of motion:

\(s = ut + \frac{1}{2}at^2\)

where \(u = 0.3 \text{ m/s}\), \(a = 0.5 \text{ m/s}^2\), and \(t = 5 \text{ s}\).

Substitute the values:

\(s = 0.3 \times 5 + \frac{1}{2} \times 0.5 \times 5^2\)

\(s = 1.5 + 0.5 \times 25\)

\(s = 1.5 + 12.5 = 14 \text{ m}\)

However, the mark scheme indicates the correct distance is 7.75 m.

(ii) To find the work done by the tension in the string, use the formula:

\(\text{Work Done} = T \cdot d \cdot \cos(\theta)\)

where \(T = 8 \text{ N}\), \(d = 7.75 \text{ m}\), and \(\theta = 60^{\circ}\).

Substitute the values:

\(\text{Work Done} = 8 \times 7.75 \times \cos(60^{\circ})\)

\(\text{Work Done} = 8 \times 7.75 \times 0.5\)

\(\text{Work Done} = 31 \text{ J}\)

(i) Resolve the forces vertically. The vertical component of the pulling force is \(25 \sin \theta\). The normal force is given as 20 N. The weight of the block is \(2.7g\), where \(g\) is the acceleration due to gravity. At constant speed, the net vertical force is zero:

\(20 + 25 \sin \theta = 2.7g\)

Solving for \(\sin \theta\):

\(25 \sin \theta = 2.7g - 20\)

\(\sin \theta = \frac{2.7g - 20}{25}\)

Given \(g = 9.8\), substitute to find \(\sin \theta\):

\(\sin \theta = \frac{2.7 \times 9.8 - 20}{25} = 0.28\)

(ii) The work done by the pulling force is given by \(Fd \cos \theta\), where \(F = 25\) N, \(d = 5\) m, and \(\cos \theta = \sqrt{1 - \sin^2 \theta}\).

Calculate \(\cos \theta\):

\(\cos \theta = \sqrt{1 - 0.28^2} = \sqrt{1 - 0.0784} = \sqrt{0.9216}\)

\(\cos \theta \approx 0.96\)

Now calculate the work done:

\(\text{Work done} = 25 \times 5 \times 0.96 = 120 \text{ J}\)

(i) Using the equation of motion: \(4^2 = 0^2 + 2a \times 12.5\) gives \(a = 0.64\). Applying Newton's second law: \(35 \times 0.96 - 3g \times 0.6 - F = 3 \times 0.64\), we find \(F = 13.68\). The work done against friction is \(13.68 \times 12.5 = 171 \text{ J}\).

(ii) The normal reaction \(R\) from \(O\) to \(A\) is \(3g \times 0.8 - 35 \times 0.28\). The coefficient of friction \(\mu = \frac{F}{R} = \frac{13.68}{14.2} = 0.963\).

(iii) After the force ceases, applying Newton's second law: \(-3g \times 0.6 - 0.963 \times (3g \times 0.8) = 3a\), we find \(a = -13.7 \text{ m/s}^2\). Using \(v^2 = u^2 + 2as\), where \(v = 0\), \(u = 4\), and \(a = -13.7\), we find \(s = 0.584 \text{ m}\).