A block of mass 'm' is placed on the floor of a lift, moving upward with an uniform acceleration a = g. (`mu`is the coefficient of friction between block and the floor of lift). If a horizontal force of umg acts on the block, then horizontal acceleration of the body is .......

Options

A. 2
B. 6
C. 8
D. Zero

Answer

To solve the problem, we need to analyze the forces acting on the block in the lift that is moving upward with an acceleration equal to \( g \). Let's break down the steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block**:
- The gravitational force acting downward: \( F_g = mg \).
- The normal force \( N \) acting upward from the lift floor.
- A pseudo force \( F_p \) acting downward due to the upward acceleration of the lift, which is equal to \( ma = mg \) (since \( a = g \)).
- A horizontal force \( F_h = \mu mg \) acting on the block.

2. **Calculate the Normal Force**:
- Since the lift is accelerating upward, the normal force \( N \) must balance both the weight of the block and the pseudo force:
\[ N = mg + F_p = mg + mg = 2mg \]

3. **Determine the Maximum Static Friction**:
- The maximum static friction \( F_s^{max} \) can be calculated using the normal force:
\[ F_s^{max} = \mu N = \mu (2mg) = 2\mu mg \]

4. **Compare the Horizontal Force with Maximum Static Friction**:
- The horizontal force acting on the block is \( F_h = \mu mg \).
- We need to check if this force exceeds the maximum static friction:
\[ F_h = \mu mg < F_s^{max} = 2\mu mg \]
- Since \( F_h \) is less than \( F_s^{max} \), the block will not slip.

5. **Determine the Net Force and Acceleration**:
- Since the block does not slip, the static friction force \( F_s \) will adjust to match the horizontal force \( F_h \):
\[ F_s = F_h = \mu mg \]
- The net force in the horizontal direction is:
\[ F_{net} = F_h - F_s = \mu mg - \mu mg = 0 \]
- Therefore, the horizontal acceleration \( a \) of the block is:
\[ a = \frac{F_{net}}{m} = \frac{0}{m} = 0 \]

### Final Answer:
The horizontal acceleration of the body is \( 0 \).

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