(i) The work done by the tension is given by the formula:

\(ext{Work Done} = T imes d imes ext{cos}( heta)\)

where \(T = 65 \text{ N}\), \(d = 76 \text{ m}\), and \(\theta = 5^{\circ}\).

\(ext{Work Done} = 65 imes 76 imes ext{cos}(5^{\circ})\)

\(ext{Work Done} = 4920 \text{ J}\)

(ii) The rate of working (power) is given by:

\(P = T imes v imes ext{cos}( heta)\)

where \(v = 1.5 \text{ m/s}\).

\(P = 65 imes 1.5 imes ext{cos}(5^{\circ})\)

\(P = 97.1 \text{ W}\)

(a) To find the power of the car's engine at point A, we first calculate the driving force using Newton's second law:

\(D - 500 = 1200 \times 0.8\)

\(D = 1460 \text{ N}\)

The power \(P\) is given by:

\(P = D \times v = 1460 \times 15 = 21900 \text{ W}\)

(b) To show that the distance AB is 1362.6 m, we first calculate the change in kinetic energy:

\(\text{Change in KE} = \frac{1}{2} \times 1200 \times 32^2 - \frac{1}{2} \times 1200 \times 15^2\)

\(= 614400 - 135000 = 479400 \text{ J}\)

The work done by the engine is:

\(\text{Work done} = 21900 \times 53 = 1160700 \text{ J}\)

Using the work-energy principle:

\(1160700 - 500 \times d = 479400\)

\(500d = 1160700 - 479400\)

\(500d = 681300\)

\(d = \frac{681300}{500} = 1362.6 \text{ m}\)

(i) The gain in potential energy (PE) is calculated as:

\(\text{Gain in PE} = mgh = 1250 \times 9.8 \times 1.54 = 19250 \text{ J}\)

The total work done by the crane is the sum of the gain in potential energy and the work done against resistance:

\(\text{Work done by crane} = 19250 + 5750 = 25000 \text{ J}\)

(ii) Using the formula for power:

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

Given \(P = 1250 \text{ W}\) and \(\Delta W = 25000 \text{ J}\), we solve for \(\Delta t\):

\(1250 = \frac{25000}{\Delta t}\)

\(\Delta t = \frac{25000}{1250} = 20 \text{ s}\)

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

\(F = ma\)

where \(m = 700 \text{ kg}\) and \(a = 2 \text{ m s}^{-2}\).

The driving force \(DF\) minus the resistance force equals the net force:

\(DF - 600 = 700 \times 2\)

\(DF - 600 = 1400\)

\(DF = 1400 + 600 = 2000 \text{ N}\)

(ii) The power \(P\) is given by:

\(P = Fv\)

where \(F = 2000 \text{ N}\) and \(v = 15 \text{ m s}^{-1}\).

\(P = 2000 \times 15 = 30000 \text{ W}\)

Thus, the rate of working is 30000 W or 30 kW.

(i) Using the formula for power, \(P = Fv\), where \(F\) is the net force, we have:

\(24000 = (R - 1250 \times 0.32) \times 20\)

Solving for \(R\), we get \(R = 800\) N.

(ii) At point \(B\), using \(P = Fv\), we have:

\(24000/29.9 - 800 = 1250a\)

Solving for \(a\), we get \(a = 0.002\) m/s².

(iii) For \(v = 30\), \(24000/30 - 800 = 0\), so the car cannot accelerate further, hence cannot reach 30 m/s.

(iv) Since \(29.9 \leq v < 30\), the speed is approximately constant.

(v) The approximately constant speed is 30 m/s with a maximum error of 0.1 m/s, or 29.95 m/s with a maximum error of 0.05 m/s, or 29.9 m/s with a maximum error of 0.1 m/s.

(vi) (a) Using \(P = \frac{\Delta WD}{\Delta t}\), \(24 = \frac{1200}{T}\), solving gives \(T = 50\) s.

(vi) (b) Using \(s = vt\), \(s = 30 \times 50\) or \(29.95 \times 50\) or \(29.9 \times 50\), gives \(s = 1500\) m or 1495 m.