A body of mass 10 kg dropped from a height 20 m, acquires a velocity of 10m/s after a falling through a distance of 20m. What is the work done by the air resistance on the body?

Options

A. 750 J
B. 1000J
C. 1500J
D. 2000J

Explanation

To solve the problem, we need to calculate the work done by air resistance on a body of mass 10 kg that is dropped from a height of 20 m and reaches a velocity of 10 m/s after falling through that distance.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the body (m) = 10 kg
- Height from which the body is dropped (h) = 20 m
- Final velocity (v) = 10 m/s
- Acceleration due to gravity (g) = 9.81 m/s² (we can approximate it to 10 m/s² for simplicity)

2. **Calculate the Potential Energy (PE) at the Height:**
\[ PE = m \cdot g \cdot h \]
Substituting the values:
\[ PE = 10 \, \text{kg} \cdot 10 \, \text{m/s}^2 \cdot 20 \, \text{m} = 2000 \, \text{J} \]

3. **Calculate the Kinetic Energy (KE) at the Final Velocity:**
\[ KE = \frac{1}{2} m v^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \cdot 10 \, \text{kg} \cdot (10 \, \text{m/s})^2 = \frac{1}{2} \cdot 10 \cdot 100 = 500 \, \text{J} \]

4. **Calculate the Work Done by Gravity (Wg):**
The work done by gravity is equal to the potential energy lost by the body as it falls:
\[ W_g = PE = 2000 \, \text{J} \]

5. **Calculate the Work Done by Air Resistance (Wa):**
The work done by air resistance can be found using the work-energy principle:
\[ W_{\text{net}} = W_g + W_a \]
Since the net work done on the body is equal to the change in kinetic energy:
\[ W_{\text{net}} = KE - 0 = KE \]
Therefore:
\[ W_a = W_{\text{net}} - W_g \]
Substituting the values:
\[ W_a = 500 \, \text{J} - 2000 \, \text{J} = -1500 \, \text{J} \]

6. **Conclusion:**
The work done by air resistance on the body is \(-1500 \, \text{J}\).

### Final Answer:
The work done by air resistance on the body is \(-1500 \, \text{J}\).