(a) The power \(P\) required to overcome the resistance is given by \(P = Fv\), where \(F = 480 \text{ N}\) and \(v = 24 \text{ m/s}\).

\(P = 480 \times 24 = 11520 \text{ W} = 11.52 \text{ kW}\).

(b) Calculate the initial kinetic energy: \(\text{KE}_{\text{before}} = \frac{1}{2} \times 1600 \times 24^2 = 460800 \text{ J}\).

Calculate the final kinetic energy: \(\text{KE}_{\text{after}} = \frac{1}{2} \times 1600 \times 32^2 = 819200 \text{ J}\).

Calculate the potential energy lost: \(\text{PE}_{\text{lost}} = 1600 \times 280 \times 0.09 = 403200 \text{ J}\).

Total work done by the engine: \(12000 \times 10 = 120000 \text{ J}\).

Work done against resistance: \(120000 + 403200 - 819200 + 460800 = 164800 \text{ J}\).

(a) Calculate the change in kinetic energy:

\(\text{KE change} = \frac{1}{2} \times 900 \times 16^2 - \frac{1}{2} \times 900 \times 11^2 = 115200 - 54450 = 60750 \text{ J}\)

Calculate the potential energy change:

\(\text{PE} = 900 \times 150 \times 0.12 = 162000 \text{ J}\)

Work done by the car's engine:

\(\text{Work done by engine} = 24000 \times 12 = 288000 \text{ J}\)

Work done against resistive forces:

\(288000 - 162000 - 115200 + 54450 = 65250 \text{ J}\)

(b) Driving force is given by:

\(\frac{32000}{v} = 1520 + 4v\)

Solving for \(v\):

\(32000 = 1520v + 4v^2\)

\(4v^2 + 1520v - 32000 = 0\)

Solving the quadratic equation gives \(v = 20 \text{ m s}^{-1}\).

(a) Using the work-energy principle, the work done by the driving force minus the work done against resistance equals the change in kinetic energy:

\(4500d - 75000 = \frac{1}{2} \times 1200 \times 25^2\)

\(4500d - 75000 = 375000\)

\(4500d = 450000\)

\(d = 100\)

(b) Using the equation of motion for car B:

\(25^2 = 0 + 2a \times 100\)

\(a = 3.125\, \text{m/s}^2\)

Using Newton's second law:

\(3200 - 1200 = m \times 3.125\)

\(m = 640\, \text{kg}\)

(c) At point P, the power is given by:

\(\text{Power} = 3200 \times 25 = 80000\, \text{W}\)

Using the formula for power:

\(\frac{80000}{v} - 1200 = 0\)

\(v = \frac{80000}{1200} = 66.7\, \text{m/s}\)

(a) The potential energy lost in 50 m is given by:

\((m + 300) g \times 50 \sin 3^{\circ}\)

Using the work-energy principle, we have:

\((m + 300) g \times 50 \sin 3^{\circ} - 40000 = 0\)

Solving for \(m\):

\(m = 1230\) kg to 3 significant figures.

Alternative method:

The resistance force \(R\) is:

\(R = \frac{40000}{50} = 800\) N

Using Newton's second law:

\((m + 300) g \sin 3^{\circ} - R = 0\)

Solving for \(m\):

\(m = 1230\) kg to 3 significant figures.

(b) Applying Newton's second law to the trailer:

\(T + 300 g \sin 3^{\circ} - 200 = 0\)

Or to the car:

\(mg \sin 3^{\circ} = T + 600\)

Solving for \(T\):

\(T = 43.0\) N to 3 significant figures.

(a) The work done against gravity is given by the potential energy formula:

\(W = mgh\)

where \(m = 600 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), and \(h = 15 \text{ m}\).

Thus, \(W = 600 \times 9.8 \times 15 = 90,000 \text{ J}\).

The total work done by the crane also includes the work done against resistance, which is 10,000 J. Therefore, the total work done is:

\(90,000 + 10,000 = 100,000 \text{ J}\).

(b) Given the average power \(P = 12.5 \text{ kW} = 12,500 \text{ W}\), we use the formula:

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

where \(W = 100,000 \text{ J}\).

Rearranging gives:

\(t = \frac{W}{P} = \frac{100,000}{12,500} = 8 \text{ s}\).