A 200 t train experiences wind resistance equivalent to \( 62.5 \mathrm{~N} / \mathrm{t} \). The operating efficiency of the driving motors is 0.87 and the cost of electrical energy is \( 0.8 \mathrm{Rand} / \mathrm{kW} \mathrm{h} \). What is the cost of the energy required to make the train travel 1 km ? If the train is supplied at a constant voltage of 1.5 kV and travels with a velocity of \( 80 \mathrm{~km} / \mathrm{h} \), what is the supply current? [R3.2; 214A]

**Step 1. Calculate the total force due to wind resistance** The wind resistance is given as \[ 62.5\ \frac{\mathrm{N}}{\mathrm{t}} \] For a \( 200\ \mathrm{t} \) train, the total resistance force is \[ F = 62.5\ \frac{\mathrm{N}}{\mathrm{t}} \times 200\ \mathrm{t} = 12500\ \mathrm{N}. \] **Step 2. Determine the work done to travel 1 km** Work done is the product of force and distance: \[ W = F \times d, \] where \( d = 1\ \mathrm{km} = 1000\ \mathrm{m} \). Thus, \[ W = 12500\ \mathrm{N} \times 1000\ \mathrm{m} = 12.5 \times 10^6\ \mathrm{J}. \] **Step 3. Adjust the work for the motor efficiency** The operating efficiency of the motors is \( \eta = 0.87 \). This means that the energy drawn from the electrical supply must be higher than the mechanical work output. The actual energy required is: \[ W_{\text{supplied}} = \frac{W}{\eta} = \frac{12.5 \times 10^6\ \mathrm{J}}{0.87} \approx 14.37 \times 10^6\ \mathrm{J}. \] **Step 4. Convert the energy to kilowatt-hours and calculate its cost** Recall that \[ 1\ \mathrm{kWh} = 3.6 \times 10^6\ \mathrm{J}. \] Thus, the energy in kilowatt-hours is: \[ E_{\mathrm{kWh}} = \frac{14.37 \times 10^6\ \mathrm{J}}{3.6 \times 10^6\ \mathrm{J/kWh}} \approx 3.99\ \mathrm{kWh}. \] The cost of electrical energy is given as \( 0.8\ \mathrm{Rand/kWh} \). Therefore, the cost is: \[ \text{Cost} = 3.99\ \mathrm{kWh} \times 0.8\ \frac{\mathrm{Rand}}{\mathrm{kWh}} \approx 3.19\ \mathrm{Rand}. \] **Step 5. Determine the supply current when the train travels at \( 80\ \mathrm{km/h} \)** 1. **Calculate the train's speed in meters per second:** \[ v = 80\ \frac{\mathrm{km}}{\mathrm{h}} = \frac{80}{3.6}\ \frac{\mathrm{m}}{\mathrm{s}} \approx 22.22\ \frac{\mathrm{m}}{\mathrm{s}}. \] 2. **Calculate the mechanical power required to overcome resistance at this speed:** \[ P_{\text{traction}} = F \times v = 12500\ \mathrm{N} \times 22.22\ \frac{\mathrm{m}}{\mathrm{s}} \approx 277777.5\ \mathrm{W}. \] 3. **Account for the driving motor efficiency:** Since the motors operate at \( 87\% \) efficiency, the electrical power drawn is: \[ P_{\text{electrical}} = \frac{P_{\text{traction}}}{0.87} \approx \frac{277777.5\ \mathrm{W}}{0.87} \approx 319080\ \mathrm{W}. \] 4. **Calculate the supply current using \( I = \frac{P}{V} \):** The supply voltage is \( 1.5\ \mathrm{kV} = 1500\ \mathrm{V} \). Therefore: \[ I = \frac{319080\ \mathrm{W}}{1500\ \mathrm{V}} \approx 212.72\ \mathrm{A}. \] Rounding, we obtain approximately \( 214\ \mathrm{A} \). **Final Answers:** - The cost of the electrical energy required to make the train travel 1 km is approximately \( 3.2\ \mathrm{Rand} \). - The supply current is approximately \( 214\ \mathrm{A} \).