A car weighing 500 kg climbs up a hill of slope 1 in 49 with a velocity of 36 KMPH. If the frictional force is 50 N, the power delivered by the engine is

Options

A. 500W
B. 5kW
C. 1.5kW
D. 150W

Answer

To find the power delivered by the engine of the car climbing up a hill, we can follow these steps:

### Step 1: Understand the problem
We have a car weighing 500 kg climbing a hill with a slope of 1 in 49 at a velocity of 36 km/h. There is a frictional force of 50 N acting against the motion of the car.

### Step 2: Convert the velocity from km/h to m/s
To work with standard units, we need to convert the velocity from kilometers per hour to meters per second:
\[ \text{Velocity} = 36 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 10 \text{ m/s} \]

### Step 3: Calculate the angle of the slope
The slope of 1 in 49 means for every 49 units of horizontal distance, there is a rise of 1 unit. We can find the sine of the angle (θ) using:
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \approx \frac{1}{49} \]
For small angles, we can approximate \(\sin \theta\) as \(\tan \theta\).

### Step 4: Calculate the gravitational force component along the slope
The force due to gravity acting down the slope can be calculated using:
\[ F_{\text{gravity}} = mg \sin \theta \]
Where:
- \(m = 500 \text{ kg}\)
- \(g = 10 \text{ m/s}^2\)
- \(\sin \theta \approx \frac{1}{49}\)

So,
\[ F_{\text{gravity}} = 500 \times 10 \times \frac{1}{49} = \frac{5000}{49} \approx 102.04 \text{ N} \]

### Step 5: Calculate the total force required to overcome gravity and friction
The total force (F) required by the engine to keep the car moving at constant velocity (overcoming both gravity and friction) is:
\[ F = F_{\text{friction}} + F_{\text{gravity}} = 50 \text{ N} + 102.04 \text{ N} = 152.04 \text{ N} \]

### Step 6: Calculate the power delivered by the engine
Power (P) delivered by the engine can be calculated using the formula:
\[ P = F \times v \]
Substituting the values we have:
\[ P = 152.04 \text{ N} \times 10 \text{ m/s} = 1520.4 \text{ W} = 1.5204 \text{ kW} \]

### Step 7: Round off the answer
Rounding off, we can say the power delivered by the engine is approximately:
\[ P \approx 1.5 \text{ kW} \]

### Final Answer:
The power delivered by the engine is approximately **1.5 kW**.

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