A lift is moving downwards with an acceleration equal to `g`. A block of mass `m` , kept on the floor of the lift of friction coefficient `mu` , is pulled horizontally . The friction acting on the block is

Options

A. `mu Mg`
B. `2 mu M g`
C. zero
D. Mg

Explanation

To solve the problem, we need to analyze the situation of the block in the lift that is accelerating downwards with an acceleration equal to \( g \).

### Step-by-Step Solution:

1. **Understanding the System**:
- The lift is moving downwards with an acceleration \( a = g \).
- A block of mass \( m \) is placed on the floor of the lift.
- The coefficient of friction between the block and the lift is \( \mu \).

2. **Reference Frame**:
- We will analyze the block from the reference frame of the lift. In this frame, the block experiences a pseudo force due to the acceleration of the lift.

3. **Forces Acting on the Block**:
- The gravitational force acting on the block is \( mg \) (downwards).
- The pseudo force acting on the block (due to the downward acceleration of the lift) is \( ma \) where \( a = g \). Thus, the pseudo force is \( mg \) (upwards).
- The normal force \( N \) acts upwards on the block from the lift floor.

4. **Net Force in the Vertical Direction**:
- Since the lift is accelerating downwards with \( g \), the net force in the vertical direction must be zero for the block to remain in equilibrium in the lift's frame.
- Therefore, we can write the equation:
\[ N + mg - mg = 0 \]
- Simplifying this gives:
\[ N = 0 \]

5. **Frictional Force Calculation**:
- The frictional force \( F_f \) acting on the block is given by the equation:
\[ F_f = \mu N \]
- Since we found that \( N = 0 \), we substitute this into the friction equation:
\[ F_f = \mu \cdot 0 = 0 \]

6. **Conclusion**:
- The friction acting on the block is zero because the normal force between the block and the lift is zero.

### Final Answer:
The friction acting on the block is \( 0 \).