(i) The power required to overcome the resistance is given by:

\(P = \text{DF} \times v = 850 \times 36\)

\(P = 30600 \text{ W} = 30.6 \text{ kW}\)

(ii) The driving force (DF) is the sum of the resistance and the component of the weight along the hill:

\(\text{DF} = 1250g \times 0.1 + 850\)

The power is given by:

\(P = \text{DF} \times v\)

\(63000 = (1250 \times 9.8 \times 0.1 + 850) \times v\)

Solving for \(v\), we get:

\(v = 30 \text{ m s}^{-1}\)

(iii) Using the energy method:

Gain in kinetic energy (KE):

\(\frac{1}{2} \times 1250 \times (24^2 - 20^2) = 110000 \text{ J}\)

Loss in potential energy (PE):

\(1250 \times 9.8 \times 176 \times 0.1 = 220000 \text{ J}\)

Work done by the car's engine:

\(20000 \times 8 = 160000 \text{ J}\)

Total work done against resistance:

\(160000 + 220000 = 270000 \text{ J} = 270 \text{ kJ}\)

(i) Using conservation of energy, the potential energy lost is converted into kinetic energy. The potential energy at A is given by:

\(mgh = 40 \times g \times 7.2\)

The kinetic energy at B is:

\(\frac{1}{2} \times 40 \times v^2\)

Equating the potential energy and kinetic energy:

\(\frac{1}{2} \times 40 \times v^2 = 40 \times g \times 7.2\)

Solving for \(v\):

\(v = 12 \text{ m s}^{-1}\)

(ii) For the rough slide, the work done against friction (WDF) is the same in both cases. First, calculate WDF for the first scenario:

\(\text{WDF} = 40 \times g \times 7.2 - \frac{1}{2} \times 40 \times 10^2 = 880\)

For the second scenario, using the work-energy principle:

\(\frac{1}{2} \times 40 \times V^2 + 40 \times g \times 7.2 = \frac{1}{2} \times 40 \times 11^2 + 880\)

Solving for \(V\):

\(V = \sqrt{21} = 4.58 \text{ m s}^{-1}\)

(i) Resolve forces perpendicular to the plane:

\(R = 0.2g \cos 30^{\circ} - T \sin 15^{\circ}\)

Frictional force \(F = \mu R = 0.3 \times (0.2g \cos 30^{\circ} - T \sin 15^{\circ})\)

Resolve forces parallel to the plane:

\(T \cos 15^{\circ} + 0.3 \times (0.2g \cos 30^{\circ} - T \sin 15^{\circ}) = 0.2g \sin 30^{\circ}\)

Solve for \(T\):

\(T = 0.541\) N

(ii) Work done against friction:

\(0.3 \times 0.2g \cos 30^{\circ} \times 3 = 1.5588\) J

Work done against 0.25 N force:

\(0.25 \times 3 = 0.75\) J

Potential energy loss:

\(0.2g \times 3 \sin 30^{\circ} = 3\) J

Using energy conservation:

\(\frac{1}{2} (0.2) v^2 = 3 - 1.5588 - 0.75\)

Solve for \(v\):

\(Speed = 2.63 m/s\)

(a) The power produced by the athlete is 60 W, and he runs at a constant speed of 3 m s^-1. The resistive force can be found using the formula for power:

\(P = F imes v\)

where \(P = 60\) W and \(v = 3\) m s^-1. Solving for \(F\):

\(F = \frac{60}{3} = 20\) N

(b) The athlete runs up a 150 m inclined section with an incline of 0.8°. The initial speed is 3 m s^-1, and the driving force is 24 N. The resistive force is 13 N. Using the work-energy principle:

Potential energy change: \(PE = 84g \times 150 \times \sin(0.8^{\circ})\)

Kinetic energy change: \(KE = \frac{1}{2} \times 84 \times v^2 - \frac{1}{2} \times 84 \times 3^2\)

Work done: \(W = (24 \times 150) - (13 \times 150)\)

Setting up the work-energy equation:

\(84g \times 150 \times \sin(0.8^{\circ}) + \frac{1}{2} \times 84 \times v^2 - \frac{1}{2} \times 84 \times 3^2 = 24 \times 150 - 13 \times 150\)

Solving for \(v\):

\(1759.23 + 42v^2 - 378 = 3600 - 1950\)

\(42v^2 = 268.765\)

\(v = 2.53\) m s^-1

(i) The work done against the resistance force is calculated using the formula:

\(W = F \times d\)

where \(F = 640 \text{ N}\) and \(d = 8 + 10 = 18 \text{ m}\).

Thus, \(W = 640 \times 18 = 11,520 \text{ J}\).

(ii) First, calculate the kinetic energy (KE) at the bottom of the first ramp:

\(\text{KE} = \frac{1}{2} \times 840 \times 14^2 = 82,320 \text{ J}\).

Next, calculate the potential energy (PE) change:

\(\text{PE gained} = 840g \times 8 \sin 30^{\circ} - 840g \times 10 \sin 20^{\circ} = 4870 \text{ J}\).

Using the work-energy principle:

\(\frac{1}{2} \times 840 \times v^2 = 82,320 - 11,520 - 4870\).

Solve for \(v\):

\(v = 12.5 \text{ m s}^{-1}\).