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 1: Identify the initial and final conditions
- The body is dropped from a height \( H \).
- Initial velocity \( u = 0 \) (since it is dropped).
- Final velocity \( v = 1.1 \sqrt{gH} \) (when it hits the ground).

### Step 2: Calculate the change in kinetic energy
The change in kinetic energy (\( \Delta KE \)) can be calculated using the formula:
\[ \Delta KE = KE_{final} - KE_{initial} \]
Where:
- \( KE_{initial} = 0 \) (since the initial velocity is 0).
- \( KE_{final} = \frac{1}{2} m v^2 = \frac{1}{2} m (1.1 \sqrt{gH})^2 \).

Calculating \( KE_{final} \):
\[ KE_{final} = \frac{1}{2} m (1.1^2 gH) = \frac{1}{2} m (1.21 gH) = 0.605 m gH \]

Thus, the change in kinetic energy is:
\[ \Delta KE = 0.605 m gH - 0 = 0.605 m gH \]

### Step 3: Calculate the work done by gravity
The work done by gravity (\( W_{gravity} \)) when the body falls from height \( H \) is given by:
\[ W_{gravity} = mgh \]

### Step 4: Apply the work-energy theorem
According to the work-energy theorem:
\[ W_{gravity} + W_{friction} = \Delta KE \]
Where \( W_{friction} \) is the work done by air friction. Rearranging gives:
\[ W_{friction} = \Delta KE - W_{gravity} \]

### Step 5: Substitute the values
Substituting the values we calculated:
\[ W_{friction} = 0.605 m gH - m gH \]
\[ W_{friction} = (0.605 - 1) m gH = -0.395 m gH \]

### Conclusion
The work done by air friction is:
\[ W_{friction} = -0.395 m gH \]

This negative sign indicates that the work done by air friction is in the opposite direction to the displacement of the body.