First, apply Newton's second law along the direction of motion:

\(F - 1130 + 1250 \cdot g \cdot \sin(3^{\circ}) = 1250 \cdot 0.2\)

where \(F\) is the driving force, \(g\) is the acceleration due to gravity (approximately 9.8 m/s²).

Rearranging gives:

\(F = 1130 - 1250 \cdot g \cdot \sin(3^{\circ}) + 250\)

Next, use the power equation:

\(P = Fv\)

Given \(P = 22000\) W, substitute for \(F\):

\(22000 = (1130 - 1250 \cdot g \cdot \sin(3^{\circ}) + 250) \cdot v\)

Solve for \(v\):

\(v = \frac{22000}{725.8} \approx 30.3 \text{ m/s}\)

(i) Using Newton's second law, the net force on the car is given by:

\(F = ma = 1200 \times 0.5 = 600 \text{ N}\)

The driving force (DF) minus the resistance is equal to the net force:

\(\text{DF} - 400 = 600\)

Thus, the driving force is:

\(\text{DF} = 1000 \text{ N}\)

The power is given by:

\(P = Fv\)

Substituting the values, we have:

\(20000 = 1000v\)

Solving for \(v\):

\(v = 20 \text{ m/s}\)

(ii) At maximum speed, the acceleration is zero, so the net force is zero:

\(\frac{20000}{v} - 400 = 0\)

Solving for \(v\):

\(\frac{20000}{v} = 400\)

\(v = 50 \text{ m/s}\)

(iii) The work done is given by:

\(\Delta W = 1500 \text{ kJ} = 1500000 \text{ J}\)

The distance \(d\) is:

\(d = \frac{1500000}{400} = 3750 \text{ m}\)

The time taken \(t\) is:

\(t = \frac{3750}{50} = 75 \text{ s}\)

To find the force produced by the engine, we use the formula for power:

\(P = F \times v\)

where \(P = 8000 \text{ W}\) (since 8 kW = 8000 W) and \(v = 25 \text{ m/s}\). Solving for \(F\):

\(F = \frac{P}{v} = \frac{8000}{25} = 320 \text{ N}\)

To find the resistance to motion, we apply Newton's second law:

\(F - R = ma\)

where \(F = 320 \text{ N}\), \(m = 100 \text{ kg}\), and \(a = 0.5 \text{ m/s}^2\). Solving for \(R\):

\(320 - R = 100 \times 0.5\)

\(320 - R = 50\)

\(R = 320 - 50 = 270 \text{ N}\)

(i) Since the crate is lifted at constant speed, the tension in the cable equals the weight of the crate. The weight is given by:

\(W = mg\)

where \(m = 800 \text{ kg}\) and \(g = 10 \text{ m/s}^2\) (standard gravity approximation). Thus,

\(W = 800 \times 10 = 8000 \text{ N}\)

Alternatively, using \(g = 9.8 \text{ m/s}^2\) or \(g = 9.81 \text{ m/s}^2\), the tension can be \(7840 \text{ N}\) or \(7850 \text{ N}\) respectively.

(ii) The power applied is calculated using the formula:

\(P = \frac{\Delta W}{\Delta t}\)

where \(\Delta W = 800 \times 20\) (work done to lift the crate by 20 m) and \(\Delta t = 50 \text{ s}\).

\(\Delta W = 16000 \text{ J}\)

\(P = \frac{16000}{50} = 320 \text{ W}\)

Alternatively, using \(g = 9.8 \text{ m/s}^2\) or \(g = 9.81 \text{ m/s}^2\), the power can be \(3140 \text{ W}\).

First, calculate the driving force using the power formula:

\(Power = Force × Velocity\)

\(Given power = 20,000 W and velocity = 25 m/s,\)

Driving force = \(\frac{20,000}{25} = 800 \text{ N}\).

Using Newton's second law, the net force is given by:

\(Net force = Driving force - Resistance = 800 - 600 = 200 \text{ N}.\)

According to Newton's second law, \(F = ma\), where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration.

Substitute the known values:

\(200 = 1000a\)

\(a = \frac{200}{1000} = 0.2 \text{ m s}^{-2}\).