A car weighing `1000 kg` is going up an incline with a slope of `2` in `25` at a steady speed of `18 kmph`. If `g = 10 ms^(-2)`, the power of its engine is

Options

A. 4 kW
B. 50 kW
C. 625 kW
D. 25 kW

Explanation

To find the power of the engine of the car going up an incline, we can follow these steps:

### Step 1: Identify the given data
- Weight of the car (m) = 1000 kg
- Slope of the incline = 2 in 25
- Speed of the car (v) = 18 km/h
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Convert the speed from km/h to m/s
To convert the speed from kilometers per hour to meters per second, we use the conversion factor:
\[ 1 \text{ km/h} = \frac{1}{3.6} \text{ m/s} \]
Thus,
\[ v = 18 \, \text{km/h} = \frac{18}{3.6} \, \text{m/s} = 5 \, \text{m/s} \]

### Step 3: Calculate the angle of the incline
The slope of 2 in 25 means that for every 25 meters horizontally, the height increases by 2 meters. The sine of the angle θ can be calculated as:
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{(2^2 + 25^2)}} = \frac{2}{\sqrt{4 + 625}} = \frac{2}{\sqrt{629}} \approx \frac{2}{25} = 0.08 \]

### Step 4: Calculate the gravitational force component along the incline
The component of the gravitational force acting down the incline is given by:
\[ F_{\text{gravity}} = mg \sin \theta \]
Substituting the values:
\[ F_{\text{gravity}} = 1000 \, \text{kg} \times 10 \, \text{m/s}^2 \times \frac{2}{25} \]
\[ F_{\text{gravity}} = 1000 \times 10 \times 0.08 = 800 \, \text{N} \]

### Step 5: Calculate the power of the engine
Power is defined as the work done per unit time, which can also be expressed as:
\[ P = F \cdot v \]
Substituting the values we found:
\[ P = 800 \, \text{N} \times 5 \, \text{m/s} = 4000 \, \text{W} \]

### Step 6: Convert power to kilowatts
To convert watts to kilowatts:
\[ P = \frac{4000 \, \text{W}}{1000} = 4 \, \text{kW} \]

### Final Answer
The power of the engine is **4 kW**.

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