A small block of mass m is kept on a rough inclined surface of inclination `theta` in an elevator.Th elevator goes up with a uniform velocity v and the block does not slide on the wedge.The work done by the force of friction on the block in time t as seen by the observer on the inclined plane will be

Options

A. zero
B. mgvt `cos^(2)theta`
C. mgvt `sin^(2)theta`
D. mgvt sin 2`theta`

Explanation

To solve the problem, we need to analyze the forces acting on the block and determine the work done by the force of friction. Let's break it down step by step:

### Step 1: Understand the System
The block of mass \( m \) is placed on a rough inclined surface with an angle of inclination \( \theta \). The elevator is moving upwards with a constant velocity \( v \). Since the elevator is moving with uniform velocity, there is no net acceleration acting on the block in the vertical direction.

### Step 2: Identify Forces Acting on the Block
1. **Weight of the Block**: The weight \( W \) of the block acts downwards and is given by \( W = mg \).
2. **Normal Force**: The normal force \( N \) acts perpendicular to the inclined surface.
3. **Frictional Force**: The force of friction \( f \) acts parallel to the inclined surface and opposes the motion of the block.

### Step 3: Analyze the Forces
Since the block does not slide on the inclined surface, it is in equilibrium relative to the inclined plane. The frictional force must balance the component of the gravitational force acting down the incline.

The component of the weight acting down the incline is given by:
\[ f = mg \sin \theta \]

### Step 4: Determine the Displacement
Since the block does not slide and is at rest relative to the inclined plane, the displacement of the block relative to the inclined plane is zero.

### Step 5: Calculate Work Done by Friction
The work done \( W \) by the frictional force is given by the formula:
\[ W = f \cdot d \cdot \cos(\phi) \]
where \( d \) is the displacement and \( \phi \) is the angle between the force and the displacement.

Since the displacement \( d = 0 \) (the block does not move), the work done by the frictional force becomes:
\[ W = f \cdot 0 \cdot \cos(\phi) = 0 \]

### Conclusion
The work done by the force of friction on the block in time \( t \) as seen by the observer on the inclined plane is:
\[ \boxed{0} \]