A train of mass 400 tons climbs up an incline of `1//49` at the rate of `36kmh^(-1)` . The force of friction is 4kg' at per ton. Find the power of the engine.

Explanation

To solve the problem, we need to find the power of the engine of a train climbing an incline. We will follow these steps:

### Step 1: Convert the mass of the train to kilograms
The mass of the train is given as 400 tons. We need to convert this to kilograms:
\[ \text{Mass} = 400 \text{ tons} = 400 \times 1000 \text{ kg} = 400000 \text{ kg} \]

### Step 2: Calculate the force of friction
The force of friction is given as 4 kg per ton. Therefore, for 400 tons:
\[ \text{Force of friction} = 4 \text{ kg/ton} \times 400 \text{ tons} = 1600 \text{ kg} \]
Now, we convert this to Newtons using \( g = 9.8 \text{ m/s}^2 \):
\[ F_s = 1600 \text{ kg} \times 9.8 \text{ m/s}^2 = 15680 \text{ N} \]

### Step 3: Calculate the gravitational force component along the incline
The incline is given as \( \frac{1}{49} \). Therefore, the sine of the angle \( \theta \) can be approximated as:
\[ \sin \theta = \frac{1}{49} \]
The gravitational force component acting down the incline is:
\[ F_g = m \cdot g \cdot \sin \theta = 400000 \text{ kg} \times 9.8 \text{ m/s}^2 \times \frac{1}{49} \]
Calculating this:
\[ F_g = 400000 \times 9.8 \times \frac{1}{49} \approx 80000 \text{ N} \]

### Step 4: Calculate the total force required to overcome friction and gravity
The total force \( F \) required by the engine to move the train up the incline is the sum of the force of friction and the gravitational force component:
\[ F = F_s + F_g = 15680 \text{ N} + 80000 \text{ N} = 95680 \text{ N} \]

### Step 5: Convert the speed of the train to meters per second
The speed of the train is given as 36 km/h. We convert this to meters per second:
\[ V = 36 \text{ km/h} = \frac{36 \times 1000}{3600} \text{ m/s} = 10 \text{ m/s} \]

### Step 6: Calculate the power of the engine
Power is calculated using the formula:
\[ \text{Power} = F \cdot V \]
Substituting the values we found:
\[ \text{Power} = 95680 \text{ N} \times 10 \text{ m/s} = 956800 \text{ W} \]
Converting this to kilowatts:
\[ \text{Power} = \frac{956800}{1000} \text{ kW} = 956.8 \text{ kW} \]

### Final Answer
The power of the engine is approximately **956.8 kW**.

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