(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 \(v\):

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