A lift moving with a uniform velocity v in which a block of mass `m` is lying .The friction force offered by the when coefficient of friction is `mu = 0.5` will be

Options

A. zero
B. `mg//2`
C. `mg`
D. `2mg`

Explanation

To solve the problem, we need to analyze the forces acting on the block of mass `m` in the lift that is moving with a uniform velocity `v`.

### Step 1: Understand the scenario
The lift is moving with a uniform velocity `v`. This means that it is not accelerating. In this case, the block of mass `m` is at rest relative to the lift.

**Hint:** Remember that uniform velocity implies no net acceleration.

### Step 2: Identify the forces acting on the block
The forces acting on the block are:
- The gravitational force acting downwards, which is `mg` (where `g` is the acceleration due to gravity).
- The normal force `N` acting upwards from the surface of the lift.

**Hint:** Draw a free-body diagram to visualize the forces acting on the block.

### Step 3: Determine the normal force
Since the lift is moving with uniform velocity, the normal force `N` will balance the gravitational force. Therefore, we have:
\[ N = mg \]

**Hint:** Use Newton's first law to understand that the net force in the vertical direction must be zero for the block to remain at rest.

### Step 4: Calculate the maximum static friction
The maximum static friction force can be calculated using the coefficient of friction `μ`:
\[ f_{\text{max}} = \mu N \]
Substituting the value of `N`:
\[ f_{\text{max}} = \mu (mg) \]
Given that `μ = 0.5`, we have:
\[ f_{\text{max}} = 0.5 \times mg \]

**Hint:** Remember that static friction adjusts to prevent relative motion up to its maximum value.

### Step 5: Analyze the motion of the block
Since the lift is moving uniformly, there is no external force trying to move the block relative to the lift. Therefore, there is no tendency for the block to slide.

**Hint:** Consider what happens when there is no applied force on the block.

### Step 6: Conclusion about the friction force
Since there is no tendency of relative motion between the block and the lift, the actual friction force acting on the block is zero:
\[ f = 0 \]

**Hint:** Recall that static friction only acts when there is an attempt to move the object.

### Final Answer
The friction force offered by the surface when the coefficient of friction is `μ = 0.5` will be **0 N**.

---