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

Options

A. zero
B. `mgvt cos^(2)theta`
C. `mgvt sin^(2)theta`
D. `(1)/(2)mgvt sin2theta`

Explanation

To solve the problem of finding the work done by the force of friction on a block resting on a rough inclined surface in an elevator moving upward with uniform velocity, we can follow these steps:

### Step 1: Understand the Forces Acting on the Block
- The block has a weight \( mg \) acting vertically downward.
- The weight can be resolved into two components:
- Perpendicular to the inclined plane: \( mg \cos \theta \)
- Parallel to the inclined plane: \( mg \sin \theta \)

### Step 2: Identify the Condition of the Block
- The block does not slide down the incline, which means that the force of friction must balance the component of gravitational force acting down the incline.
- Therefore, the force of friction \( f \) is equal to \( mg \sin \theta \).

### Step 3: Determine the Work Done by the Force of Friction
- Work done \( W \) by the force of friction is given by the formula:
\[ W = f \cdot s \cdot \cos(\phi) \]
where:
- \( f \) is the force of friction,
- \( s \) is the displacement of the block,
- \( \phi \) is the angle between the force of friction and the direction of displacement.

### Step 4: Calculate the Displacement
- Since the elevator moves with a uniform velocity \( v \), the displacement \( s \) of the block in time \( t \) is:
\[ s = vt \]

### Step 5: Determine the Angle Between Forces
- The angle \( \phi \) between the direction of the force of friction (down the incline) and the displacement (up the incline) is \( 90^\circ - \theta \).
- Therefore, \( \cos(\phi) = \cos(90^\circ - \theta) = \sin(\theta) \).

### Step 6: Substitute Values into the Work Done Formula
- Substitute \( f = mg \sin \theta \), \( s = vt \), and \( \cos(\phi) = \sin(\theta) \) into the work done formula:
\[ W = (mg \sin \theta) \cdot (vt) \cdot \sin(\theta) \]

### Step 7: Simplify the Expression
- Simplifying the expression gives:
\[ W = mg vt \sin^2 \theta \]

### Final Answer
- The work done by the force of friction on the block in time \( t \) is:
\[ W = mg vt \sin^2 \theta \]

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