A body dropped from a height 'H' reaches the ground with a speed of `1.1 sqrt(gH)`. Calculate the work done by air friction :

Options

A. 0.395 mgH
B. `-0.395` mgH
C. 0.400 mgH
D. `-0.400` mgH

Answer

To solve the problem of calculating the work done by air friction on a body dropped from a height \( H \) that reaches the ground with a speed of \( 1.1 \sqrt{gH} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial and Final Conditions:**
- The body is dropped from a height \( H \).
- Initial velocity \( u = 0 \) (since it is dropped).
- Final velocity \( v = 1.1 \sqrt{gH} \).

2. **Use the Work-Energy Theorem:**
- According to the work-energy theorem, the total work done on the body is equal to the change in kinetic energy:
\[ W_{\text{total}} = \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \]
- The initial kinetic energy \( KE_{\text{initial}} = 0 \) (since \( u = 0 \)).
- The final kinetic energy can be calculated as:
\[ KE_{\text{final}} = \frac{1}{2} m v^2 = \frac{1}{2} m (1.1 \sqrt{gH})^2 = \frac{1}{2} m (1.21 gH) = 0.605 mgH \]
- Therefore, the total work done is:
\[ W_{\text{total}} = 0.605 mgH - 0 = 0.605 mgH \]

3. **Calculate Work Done by Gravity:**
- The work done by gravity when the body falls a height \( H \) is given by:
\[ W_{\text{gravity}} = mgh \]
- Since we are taking downward as positive, this work is:
\[ W_{\text{gravity}} = mgH \]

4. **Calculate Work Done by Air Friction:**
- The work done by air friction can be found using the relation:
\[ W_{\text{friction}} = W_{\text{total}} - W_{\text{gravity}} \]
- Substituting the values we calculated:
\[ W_{\text{friction}} = 0.605 mgH - mgH = (0.605 - 1) mgH = -0.395 mgH \]

5. **Final Answer:**
- The work done by air friction is:
\[ W_{\text{friction}} = -0.395 mgH \]