A block is gently placed on a conveyor belt moving horizontally with constant speed. After t = 4s, the velocity of the block becomes equal to the velocity of the belt. If the coefficient of friction between the block and the belt is `mu = 0.2`, then the velocity of the conveyor belt is `(g= 10ms^(-2)) `

Options

A. `2 ms^(-1)`
B. `4 ms^(-1)`
C. `64ms^(-1)`
D. `8 ms^(-1)`

Answer

To solve the problem, we need to find the velocity of the conveyor belt (v) given the time (t = 4 seconds) and the coefficient of friction (μ = 0.2).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A block is placed on a conveyor belt moving at a constant speed. The block starts from rest and accelerates due to friction until its velocity matches that of the conveyor belt after 4 seconds.

2. **Identify the Forces**:
- The force of friction (F_friction) acting on the block is what causes it to accelerate. This force can be expressed as:
\[ F_{\text{friction}} = \mu \cdot m \cdot g \]
- Where:
- \( \mu = 0.2 \) (coefficient of friction)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( m \) is the mass of the block (which will cancel out later).

3. **Calculate the Frictional Force**:
- Substituting the values into the equation:
\[ F_{\text{friction}} = 0.2 \cdot m \cdot 10 = 2m \, \text{N} \]

4. **Using Newton's Second Law**:
- According to Newton's second law, the acceleration (a) of the block can be expressed as:
\[ F_{\text{friction}} = m \cdot a \]
- Therefore, we have:
\[ 2m = m \cdot a \implies a = 2 \, \text{m/s}^2 \]

5. **Finding the Final Velocity**:
- The final velocity (v) of the block after time \( t \) can be calculated using the equation of motion:
\[ v = u + a \cdot t \]
- Since the initial velocity \( u = 0 \):
\[ v = 0 + 2 \cdot 4 = 8 \, \text{m/s} \]

6. **Conclusion**:
- The velocity of the conveyor belt, which is the final velocity of the block after 4 seconds, is:
\[ v = 8 \, \text{m/s} \]

### Final Answer:
The velocity of the conveyor belt is \( 8 \, \text{m/s} \).