An engine of one metric ton is going up an inclined plane, 1 in 2 at the rate of 36 kmph. If the coefficient of friction is `1//sqrt(3)` , the power of engine is

Options

A. 9.8 W
B. 98 W
C. 980 W
D. 98 kW

Answer

To solve the problem, we need to calculate the power of the engine that is moving up an inclined plane. Let's break it down step by step.

### Step 1: Convert the mass of the engine to kilograms
The mass of the engine is given as 1 metric ton.
1 metric ton = 1000 kg.

### Step 2: Convert the speed from km/h to m/s
The speed of the engine is given as 36 km/h.
To convert km/h to m/s, we use the conversion factor:
\[ \text{Speed (m/s)} = \text{Speed (km/h)} \times \frac{5}{18} \]
So,
\[ \text{Speed} = 36 \times \frac{5}{18} = 10 \text{ m/s} \]

### Step 3: Determine the angle of inclination
The incline is given as 1 in 2, which means for every 1 unit of height, there are 2 units of horizontal distance.
Using trigonometry:
- The height (opposite side) = 1
- The base (adjacent side) = 2
- The hypotenuse can be calculated using Pythagoras theorem:
\[ \text{Hypotenuse} = \sqrt{1^2 + 2^2} = \sqrt{5} \]
- Therefore,
\[ \sin \theta = \frac{1}{\sqrt{5}} \]
\[ \cos \theta = \frac{2}{\sqrt{5}} \]

### Step 4: Calculate the gravitational force components
The weight of the engine (force due to gravity) is:
\[ F_g = mg = 1000 \times 9.8 = 9800 \text{ N} \]
Now, we can find the components of the gravitational force:
- The component along the incline (downwards):
\[ F_{\text{gravity, parallel}} = mg \sin \theta = 9800 \times \frac{1}{\sqrt{5}} \]
- The component perpendicular to the incline (normal force):
\[ F_{\text{gravity, perpendicular}} = mg \cos \theta = 9800 \times \frac{2}{\sqrt{5}} \]

### Step 5: Calculate the frictional force
The frictional force (F_f) is given by:
\[ F_f = \mu \times F_{\text{normal}} \]
Where \( \mu = \frac{1}{\sqrt{3}} \) and \( F_{\text{normal}} = mg \cos \theta \):
\[ F_f = \frac{1}{\sqrt{3}} \times (9800 \times \frac{2}{\sqrt{5}}) \]

### Step 6: Calculate the total force required to move the engine up the incline
The total force (F_total) required to move the engine up the incline is the sum of the gravitational force component along the incline and the frictional force:
\[ F_{\text{total}} = F_{\text{gravity, parallel}} + F_f \]

### Step 7: Calculate the power of the engine
Power (P) is given by the formula:
\[ P = F_{\text{total}} \times v \]
Where \( v \) is the speed in m/s.

### Final Calculation
Now we can substitute the values we have calculated into the equations to find the power of the engine.