(i) When the lorry is going up the hill at a constant speed, the force required is given by:

\(F = 1480 + 7850g \sin 3^{\circ}\)

Calculating the force:

\(F = 1480 + 7850 \times 9.8 \times \sin 3^{\circ} = 5588 \text{ N}\)

The power is given by:

\(P = Fv = 5588 \times 10 = 55,880 \text{ W}\)

Rounding to three significant figures, the power is 55,900 W.

(ii) When the lorry is going down the hill with acceleration, use Newton's Second Law:

\(F + 7850g \sin 3^{\circ} - 1480 = 7850 \times 0.8\)

Solving for the force:

\(F = 3652 \text{ N}\)

The power is given by:

\(P = Fv = 3652 \times 15 = 54,780 \text{ W}\)

Rounding to three significant figures, the power is 54,800 W.

(i) To find the power output of the cyclist, we first calculate the resistance force: \(F_{\text{resistance}} = \frac{600}{25} = 24 \text{ N}\).

The weight component along the incline is \(80g \sin \alpha = 80 \times 9.8 \times 0.04 = 32 \text{ N}\).

The total force exerted by the cyclist is \(24 + 32 = 56 \text{ N}\).

Power is given by \(P = Fv\), where \(F\) is the total force and \(v\) is the velocity. Thus, \(P = 56 \times 4 = 224 \text{ W}\).

(ii) The work done by the cyclist in 10 seconds is \(224 \times 10 = 2240 \text{ J}\).

The initial kinetic energy (KE) at the top of the hill is \(\frac{1}{2} \times 80 \times 4^2 = 640 \text{ J}\).

Using the work-energy principle: \(\frac{1}{2} \times 80 \times v^2 = 640 + 2240 - 1200\).

Solving for \(v\), we get \(v^2 = 42\), so \(v = \sqrt{42} \approx 6.48 \text{ m s}^{-1}\).

(i) The power output of the tractor’s engine is given by the formula:

\(P = F \times v\)

where \(F = 1150 \text{ N}\) and \(v = 12 \text{ m s}^{-1}\).

Thus, \(P = 1150 \times 12 = 13,800 \text{ W}\).

(ii) The driving force when the power is increased to 25 kW is:

\(F = \frac{25000}{12}\)

Using Newton’s second law up the slope:

\(\frac{25000}{12} - 1150 - 3700g \sin 4^{\circ} = 3700a\)

Solving for \(a\):

\(a = -0.445 \text{ m s}^{-2}\)

(iii) For constant speed \(v\) on the hill:

\(\frac{25000}{v} - 1150 - 3700g \sin 4^{\circ} = 0\)

Solving for \(v\):

\(v = 6.70 \text{ m s}^{-1}\)

(i)(a) Using the formula for power, \(P = Fv\), where \(P = 16000\) W and \(v = 40\) m/s, we have:

\(16000 = F \times 40\)

Solving for \(F\), we get \(F = 400\) N.

(i)(b) With the increased power, \(P = 22500\) W and \(v = 45\) m/s, we find the force:

\(22500 = F \times 45\)

\(F = 500\) N.

Using Newton's Second Law, \(F - ext{resistance} = ma\):

\(500 - 400 = 1200a\)

\(a = \frac{1}{12} = 0.0833 \text{ m/s}^2\)

(ii) Given \(16000 = (590 + 2v)v\), solve for \(v\):

\(16000 = 590v + 2v^2\)

\(2v^2 + 590v - 16000 = 0\)

Solving this quadratic equation gives \(v = 25 \text{ m/s}\).

(i)(a) The power developed by the engine is given by the formula:

\(P = Fv\)

where \(F = 850 \text{ N}\) and \(v = 42 \text{ m s}^{-1}\).

Thus, \(P = 850 \times 42 = 35700 \text{ W} = 35.7 \text{ kW}\).

(i)(b) The new power is \(35700 + 6000 = 41700 \text{ W}\).

The driving force \(DF\) is given by:

\(DF = \frac{41700}{42} = 993 \text{ N}\).

Using Newton's second law:

\(993 - 850 = 1200a\)

\(143 = 1200a\)

\(a = \frac{143}{1200} = 0.119 \text{ m s}^{-2}\).

(ii) The driving force \(DF\) is given by:

\(DF = \frac{80000}{24} = 3333.33 \text{ N}\).

Using Newton's second law along the incline:

\(DF - 850 - 1200g \sin \theta = 0\)

\(3333.33 - 850 = 1200 \times 9.8 \times \sin \theta\)

\(2483.33 = 11760 \sin \theta\)

\(\sin \theta = \frac{2483.33}{11760}\)

\(\theta = \sin^{-1} \left( \frac{2483.33}{11760} \right) \approx 11.9^\circ\).