A block is gently placed on a conveyor belt moving horizontal with constant speed After `t = 4s` the velocity of the block becomes equal to 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 .

Options

A. `8 ms^(-1)`
B. `4 ms^(-1)`
C. `6 ms^(-1)`
D. `18 ms^(-1)`

Explanation

To solve the problem, we need to analyze the motion of the block placed on the conveyor belt. Here’s a step-by-step breakdown of the solution:

### Step 1: Understanding the Problem
The block is placed on a conveyor belt that is moving with a constant speed. Initially, the block is at rest relative to the ground. The block starts to accelerate due to the friction between it and the conveyor belt until its velocity matches that of the belt after a time `t = 4 seconds`.

### Step 2: Identify the Forces Acting on the Block
The only horizontal force acting on the block is the frictional force, which provides the necessary acceleration for the block to catch up with the conveyor belt. The frictional force can be expressed as:
\[ F_f = \mu \cdot m \cdot g \]
where:
- \( \mu = 0.2 \) (coefficient of friction)
- \( m \) is the mass of the block (which we will see cancels out)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

### Step 3: Calculate the Acceleration of the Block
The acceleration \( a \) of the block can be calculated from the frictional force:
\[ a = \frac{F_f}{m} = \frac{\mu \cdot m \cdot g}{m} = \mu \cdot g \]
Substituting the values:
\[ a = 0.2 \cdot 10 = 2 \, \text{m/s}^2 \]

### Step 4: Use the Kinematic Equation
Since the block starts from rest, we can use the kinematic equation to find the final velocity \( v \) of the block after time \( t = 4 \, \text{s} \):
\[ v = u + at \]
where:
- \( u = 0 \) (initial velocity of the block)
- \( a = 2 \, \text{m/s}^2 \)
- \( t = 4 \, \text{s} \)

Substituting the values:
\[ v = 0 + (2 \cdot 4) = 8 \, \text{m/s} \]

### Step 5: Conclusion
The final velocity of the block after 4 seconds is equal to the velocity of the conveyor belt, which is:
\[ v = 8 \, \text{m/s} \]

Thus, the velocity of the conveyor belt is **8 m/s**.

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