A lift is moving down with an acceleration equal to the acceleration due to gravity. A body of mass `M` kept on the floor of the lift is pulled horizontally If the coefficient of friction is `mu` then the frictional resistance offered by the body is .

Options

A. `mu_(k)Mg`
B. `Mg`
C. Zero
D. `mu_(k)Mg^(2)`

Answer

To solve the problem, we need to analyze the forces acting on the body of mass \( M \) placed on the floor of the lift that is accelerating downwards with an acceleration equal to the acceleration due to gravity \( g \).

### Step 1: Identify the forces acting on the body
- The gravitational force acting on the body is given by:
\[ F_g = M \cdot g \]
- The lift is accelerating downwards with \( g \), which means the effective acceleration acting on the body is zero in the vertical direction. This is because the downward acceleration of the lift cancels out the gravitational pull on the body.

### Step 2: Determine the normal force
- Since the lift is accelerating downwards with the same acceleration as gravity, the normal force \( N \) acting on the body is:
\[ N = 0 \]
- This is because the body is in a state of free fall relative to the lift.

### Step 3: Calculate the frictional force
- The frictional force \( F_f \) can be calculated using the formula:
\[ F_f = \mu \cdot N \]
- Since we found that \( N = 0 \), substituting this into the equation gives:
\[ F_f = \mu \cdot 0 = 0 \]

### Conclusion
- The frictional resistance offered by the body is:
\[ F_f = 0 \]