A coolie 1.5 m tall raises a load of 80 kg in 2 sec from the ground to his head and then walks a distance of 40 m in another 2 second. The power developed by the coolie is : `(g = 10 m//s^(2))`

Options

A. 0.2 kW
B. 0.4 kW
C. 0.6 kW
D. 0.8 kW

Answer

To solve the problem, we need to calculate the power developed by the coolie while raising the load and walking. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the Work Done in Lifting the Load
The work done (W) in lifting the load can be calculated using the formula:
\[ W = F \cdot d \]
where:
- \( F \) is the force applied (which is equal to the weight of the load),
- \( d \) is the displacement (height raised).

The force \( F \) can be calculated as:
\[ F = m \cdot g \]
where:
- \( m = 80 \, \text{kg} \) (mass of the load),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).

Now, substituting the values:
\[ F = 80 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 800 \, \text{N} \]

The displacement \( d \) is the height of the coolie:
\[ d = 1.5 \, \text{m} \]

Now, we can calculate the work done:
\[ W = 800 \, \text{N} \cdot 1.5 \, \text{m} = 1200 \, \text{J} \]

### Step 2: Calculate the Total Time Taken
The total time taken for both lifting and walking is:
\[ t = 2 \, \text{s} + 2 \, \text{s} = 4 \, \text{s} \]

### Step 3: Calculate the Power Developed
Power (P) is defined as the work done per unit time:
\[ P = \frac{W}{t} \]

Substituting the values we have:
\[ P = \frac{1200 \, \text{J}}{4 \, \text{s}} = 300 \, \text{W} \]

### Step 4: Convert Power to Kilowatts
Since the options are given in kilowatts, we need to convert watts to kilowatts:
\[ P = \frac{300 \, \text{W}}{1000} = 0.3 \, \text{kW} \]

### Conclusion
The power developed by the coolie is:
\[ \text{Power} = 0.3 \, \text{kW} \]

### Final Answer
The correct answer is not among the given options, but based on our calculations, the power developed by the coolie is 0.3 kW.