A small block of mass `2kg` is kept on a rough inclined surface of inclination `theta=30^@` fixed in a lift. The lift goes up with a uniform speed of `1ms^-1` and the block does not slide relative to the inclined surface. The work done by the force of friction on the block in a time interval of `2s` is

Options

A. (a) Zero
B. (b) `9.8J`
C. (c) `29.4J`
D. (d) `16.9J`

Explanation

To solve the problem step by step, we will analyze the forces acting on the block and calculate the work done by the friction force.

### Step 1: Identify the forces acting on the block
The block has a weight \( W = mg \) acting downwards, where \( m = 2 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). The components of this weight can be resolved into two directions:
- Perpendicular to the inclined plane: \( W_{\perp} = mg \cos(\theta) \)
- Parallel to the inclined plane: \( W_{\parallel} = mg \sin(\theta) \)

### Step 2: Calculate the weight components
Given \( \theta = 30^\circ \):
- \( W_{\parallel} = mg \sin(30^\circ) = 2 \cdot 9.8 \cdot \frac{1}{2} = 9.8 \, \text{N} \)
- \( W_{\perp} = mg \cos(30^\circ) = 2 \cdot 9.8 \cdot \frac{\sqrt{3}}{2} = 9.8 \sqrt{3} \, \text{N} \)

### Step 3: Analyze the motion of the block
Since the lift moves upward with a uniform speed of \( 1 \, \text{m/s} \), the block does not slide down the incline. This means that the net force along the incline is zero. Therefore, the frictional force must balance the component of the gravitational force acting down the incline.

### Step 4: Determine the frictional force
The frictional force \( F_f \) must equal the gravitational force component along the incline:
\[ F_f = mg \sin(30^\circ) = 9.8 \, \text{N} \]

### Step 5: Calculate the displacement of the block
The lift moves with a uniform speed of \( 1 \, \text{m/s} \) for a time interval of \( 2 \, \text{s} \):
\[ \text{Displacement} = \text{velocity} \times \text{time} = 1 \, \text{m/s} \times 2 \, \text{s} = 2 \, \text{m} \]

### Step 6: Calculate the work done by the friction force
The angle between the friction force (acting up the incline) and the displacement (along the incline) is \( 60^\circ \) (since the incline is at \( 30^\circ \) and the lift moves vertically). The work done by the friction force is given by:
\[ W = F_f \cdot d \cdot \cos(60^\circ) \]
Substituting the values:
\[ W = 9.8 \, \text{N} \cdot 2 \, \text{m} \cdot \frac{1}{2} = 9.8 \, \text{J} \]

### Final Answer
The work done by the force of friction on the block in a time interval of \( 2 \, \text{s} \) is \( 9.8 \, \text{J} \).

---