(a) Apply Newton's second law: The net force on the car is given by the equation:

\(DF - 650 = 1800 \times 0.5\)

Solving for DF gives:

\(DF = 1800 \times 0.5 + 650 = 1550\)

The power P is given by:

\(\frac{P}{20} - 650 = 1800 \times 0.5\)

Solving for P gives:

\(P = 1550 \times 20 = 31,000 ext{ W or } 31 ext{ kW}\)

(b) To find the steady speed, use the power equation:

\(\frac{31,000}{v} - 650 = 0\)

Solving for v gives:

\(v = \frac{31,000}{650} = 47.7 ext{ m s}^{-1}\)

(a) The power \(P\) is given by the formula \(P = Fv\), where \(F\) is the force and \(v\) is the velocity.

Here, \(F = 350 \text{ N}\) and \(v = 20 \text{ m s}^{-1}\).

Thus, \(P = 350 \times 20 = 7000 \text{ W} = 7 \text{ kW}\).

(b) The power \(P\) is 15 kW, which is \(15000 \text{ W}\).

Using \(P = Fv\), we have \(15000 = DF \times 20\), where \(DF\) is the driving force.

Solving for \(DF\), we get \(DF = 750 \text{ N}\).

Using Newton's second law, \(DF - 350 = 1400a\).

Substituting \(DF = 750\), we have \(750 - 350 = 1400a\).

Thus, \(400 = 1400a\).

Solving for \(a\), we get \(a = \frac{400}{1400} = \frac{2}{7} \approx 0.286 \text{ m s}^{-2}\).

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

\(F - 900 = 4000 imes 0.5\)

Solving for the driving force, F:

\(F = 900 + 4000 imes 0.5 = 2900 ext{ N}\)

(b) The power required to maintain a constant speed is given by:

\(P = F imes v\)

Here, the force F is the resistance, 900 N, and the speed v is 25 m/s:

\(P = 900 imes 25 = 22500 ext{ W} ext{ or } 22.5 ext{ kW}\)

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

\(P = F \times v\)

where \(F = 750 \text{ N}\) and \(v = 32 \text{ m/s}\).

\(P = 750 \times 32 = 24000 \text{ W} = 24 \text{ kW}\).

(a)(ii) The initial power is 24 kW, and it is decreased by 8 kW, so the new power is:

\(16000 \text{ W} = DF \times 32\)

\(DF = 500 \text{ N}\)

The net force is:

\(500 - 750 = 1250 \times a\)

\(a = -0.2 \text{ m/s}^2\)

(b) The resistance to motion is given by:

\(DF = 1000 + 8v + 1250 \times 10 \times 0.096\)

\(DF = 2200 + 8v\)

The power equation is:

\(60000 = (2200 + 8v) \times v\)

\(8v^2 + 2200v - 60000 = 0\)

Solving this quadratic equation gives:

\(v = 25 \text{ m/s}\)

(a) To find the tension \(T\) in the tow-bar, consider the forces acting on the trailer. The trailer has a mass of 400 kg and is accelerating at 1.5 \(\text{ms}^{-2}\). The net force on the trailer is given by Newton's second law:

\(T - 100 = 400 \times 1.5\)

Solving for \(T\):

\(T = 400 \times 1.5 + 100 = 700 \text{ N}\)

(b) To find the power of the engine, first calculate the total force \(F\) exerted by the engine. The car and trailer together have a total mass of 2200 kg. Using Newton's second law for the entire system:

\(F - 250 - 100 = 2200 \times 1.5\)

\(F = 2200 \times 1.5 + 250 + 100 = 3650 \text{ N}\)

The power \(P\) is given by the product of the force and the velocity:

\(P = Fv = 3650 \times 20 = 73,000 \text{ W}\)

Therefore, the power of the engine is 73,000 W or 73 kW.