A body of mass m kept on the floor of a lift moving downwards is pulled horizontally. If `mu` is the coefficient of friction between the surface in contact, then

Options

A. frictional resistance offered by the floor is 2 `mumg`, when lift moves up with a uniform velocity of 5 m/s.
B. rictional resistance offered by the floor is `muumg`, when lift moves lift moves up with uniform velocity of 3 m/s.
C. frictional resistance offered by the floor is `4.8 mumg`, when lift accelerates down with an acceleration of `4.8 m//s^(2)`
D. frictional resistance offered by the floor must lie in the range `0 le f lt oo`

Answer

To solve the problem, we need to analyze the forces acting on a body of mass \( m \) placed on the floor of a lift that is moving downwards. We will consider different scenarios based on the motion of the lift and determine the frictional force in each case.

### Step-by-Step Solution:

1. **Understanding the Scenario**:
- A body of mass \( m \) is placed on the floor of a lift.
- The lift is moving downwards, and we need to analyze the frictional force when the body is pulled horizontally.

2. **Identify Forces Acting on the Body**:
- The weight of the body acting downwards: \( W = mg \).
- The normal force \( N \) acting upwards, which will be affected by the acceleration of the lift.

3. **Case 1: Lift Moving Down with Uniform Velocity**:
- When the lift moves down with uniform velocity, the acceleration \( a = 0 \).
- The normal force \( N \) is equal to the weight of the body:
\[ N = mg \]
- The maximum frictional force \( F_r \) is given by:
\[ F_r = \mu N = \mu mg \]

4. **Case 2: Lift Accelerating Downwards**:
- When the lift accelerates downwards with acceleration \( a \), the effective weight of the body is reduced.
- The normal force \( N \) is given by:
\[ N = mg - ma = m(g - a) \]
- The frictional force is then:
\[ F_r = \mu N = \mu m(g - a) \]

5. **Case 3: Lift Moving Up with Uniform Velocity**:
- Similar to the first case, when the lift moves up with uniform velocity, the acceleration \( a = 0 \).
- The normal force remains:
\[ N = mg \]
- Thus, the frictional force is:
\[ F_r = \mu mg \]

6. **Conclusion**:
- The frictional force when the lift is moving down with uniform velocity is \( \mu mg \).
- The frictional force when the lift is accelerating downwards is \( \mu m(g - a) \).
- The frictional force when the lift is moving up with uniform velocity is also \( \mu mg \).

### Summary of Results:
- **Frictional Force when Lift Moves Down with Uniform Velocity**: \( F_r = \mu mg \)
- **Frictional Force when Lift Accelerates Downwards**: \( F_r = \mu m(g - a) \)
- **Frictional Force when Lift Moves Up with Uniform Velocity**: \( F_r = \mu mg \)