A truck of mass 30,000 kg moves up an inclined plane of slope 1 in 100 at a speed of 30 kmph . The power of the truck is `(g=10ms^(-1))`

Options

A. 25 kW
B. 10 kW
C. 5 kW
D. 2.5 kW

Answer

To find the power of the truck moving up the inclined plane, we can follow these steps:

### Step 1: Convert the speed from km/h to m/s
The speed of the truck is given as 30 km/h. We need to convert this to meters per second (m/s).

\[ \text{Speed in m/s} = \frac{30 \text{ km/h} \times 1000 \text{ m/km}}{3600 \text{ s/h}} = \frac{30000 \text{ m}}{3600 \text{ s}} \approx 8.33 \text{ m/s} \]

### Step 2: Determine the height gained per unit distance
The slope of the inclined plane is given as 1 in 100. This means for every 100 units of horizontal distance, the height gained is 1 unit.

\[ \text{Height gained (H)} = \frac{1}{100} \times \text{Distance (D)} \]

### Step 3: Calculate the gravitational potential energy change
The change in gravitational potential energy (U) when the truck moves up a height \(H\) is given by:

\[ U = mgh \]

Where:
- \(m = 30000 \text{ kg}\) (mass of the truck)
- \(g = 10 \text{ m/s}^2\) (acceleration due to gravity)
- \(h = \frac{1}{100}D\) (height gained)

### Step 4: Calculate the power
Power (P) is defined as the rate of doing work or the rate of change of energy. The power required to lift the truck can be calculated using the formula:

\[ P = \frac{U}{t} = \frac{mgh}{t} \]

Where \(t\) is the time taken to travel distance \(D\). The time \(t\) can be calculated using the formula:

\[ t = \frac{D}{v} \]

Substituting \(h\) in terms of \(D\):

\[ P = \frac{mg \left(\frac{1}{100}D\right)}{\frac{D}{v}} = mg \cdot \frac{v}{100} \]

### Step 5: Substitute the values
Now, substituting the known values:

\[ P = 30000 \text{ kg} \times 10 \text{ m/s}^2 \times \frac{8.33 \text{ m/s}}{100} \]

Calculating this:

\[ P = 30000 \times 10 \times 0.0833 = 24990 \text{ W} \approx 25 \text{ kW} \]

### Final Answer
The power of the truck is approximately **25 kW**.

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