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]

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1. **Determine the total wind resistance force on the train**  
   Given that the train weighs 200 t and the resistance is  
   $$
   62.5\ \frac{\mathrm{N}}{\mathrm{t}},
   $$  
   the total force is  
   $$
   F = 62.5\ \frac{\mathrm{N}}{\mathrm{t}} \times 200\ \mathrm{t} = 12500\ \mathrm{N}.
   $$

2. **Calculate the mechanical work required to travel 1 km**  
   The work needed to overcome this force over a distance of 1 km (i.e. $$1000\ \mathrm{m}$$) is  
   $$
   W_{\text{mech}} = F \times d = 12500\ \mathrm{N} \times 1000\ \mathrm{m} = 1.25\times10^7\ \mathrm{J}.
   $$

3. **Include the efficiency of the driving motors**  
   With an operating efficiency of 0.87, the energy drawn from the electrical supply must be higher. Thus, the electrical energy required is  
   $$
   E_{\text{elec}} = \frac{W_{\text{mech}}}{0.87} = \frac{1.25\times10^7\ \mathrm{J}}{0.87} \approx 1.4368\times10^7\ \mathrm{J}.
   $$

4. **Convert the energy to kilowatt-hours (kWh)**  
   Since  
   $$
   1\ \mathrm{kWh} = 3.6\times10^6\ \mathrm{J},
   $$  
   we have  
   $$
   E_{\text{kWh}} = \frac{1.4368\times10^7\ \mathrm{J}}{3.6\times10^6\ \mathrm{J/kWh}} \approx 3.99\ \mathrm{kWh}.
   $$

5. **Determine the cost of the electrical energy per kilometer**  
   With the cost of electrical energy given as  
   $$
   0.8\ \frac{\mathrm{Rand}}{\mathrm{kWh}},
   $$  
   the cost to travel 1 km is  
   $$
   \text{Cost} = 3.99\ \mathrm{kWh} \times 0.8\ \frac{\mathrm{Rand}}{\mathrm{kWh}} \approx 3.19\ \mathrm{Rand}.
   $$

6. **Calculate the supply current when traveling at 80 km/h**  
   First, convert 80 km/h to meters per second:  
   $$
   v = \frac{80\times1000\ \mathrm{m}}{3600\ \mathrm{s}} \approx 22.22\ \mathrm{m/s}.
   $$

The mechanical power needed to overcome wind resistance is  
   $$
   P_{\text{mech}} = F \times v = 12500\ \mathrm{N} \times 22.22\ \mathrm{m/s} \approx 277777.5\ \mathrm{W}.
   $$

Taking into account the efficiency of 0.87, the electrical power supplied must be  
   $$
   P_{\text{elec}} = \frac{P_{\text{mech}}}{0.87} \approx \frac{277777.5\ \mathrm{W}}{0.87} \approx 319000\ \mathrm{W}.
   $$

7. **Find the supply current given a constant voltage of 1.5 kV**  
   With the supply voltage  
   $$
   V = 1500\ \mathrm{V},
   $$  
   the current is given by  
   $$
   I = \frac{P_{\text{elec}}}{V} \approx \frac{319000\ \mathrm{W}}{1500\ \mathrm{V}} \approx 212.67\ \mathrm{A}.
   $$

Rounding appropriately, we have  
   $$
   I \approx 214\ \mathrm{A}.
   $$

**Final Answers:**  
- The cost of the energy required to make the train travel 1 km is approximately $$ 3.19\ \mathrm{Rand} $$.  
- The supply current is approximately $$ 214\ \mathrm{A} $$.
The cost to make the train travel 1 km is about 3.19 Rand. The supply current is approximately 214 A.

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