A conveyor belt is moving at a constant speed of `2m//s` . A box is gently dropped on it. The coefficient of friction between them is `mu=0.5` . The distance that the box will move relative to belt before coming to rest on it taking `g=10ms^(-2)` is:

Options

A. 0.4m
B. 1.2m
C. 0.6m
D. 0

Explanation

To solve the problem, we need to determine the distance that the box will move relative to the conveyor belt before coming to rest. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Problem
The box is dropped onto a conveyor belt that is moving at a constant speed of 2 m/s. The coefficient of friction (μ) between the box and the belt is 0.5. We need to find out how far the box will slide on the belt before it stops relative to the belt.

### Step 2: Identify Initial Conditions
- Initial speed of the box (u) = 2 m/s (since it was dropped onto the moving belt).
- Final speed of the box (v) = 0 m/s (when it comes to rest relative to the belt).
- Coefficient of friction (μ) = 0.5.
- Acceleration due to gravity (g) = 10 m/s².

### Step 3: Calculate the Frictional Force
The frictional force (f) acting on the box can be calculated using:
\[ f = \mu \cdot m \cdot g \]
where \( m \) is the mass of the box. However, since we are looking for acceleration, we can express it as:
\[ f = \mu \cdot m \cdot g \]

### Step 4: Calculate the Acceleration
The acceleration (a) of the box due to friction can be found using Newton's second law:
\[ a = \frac{f}{m} = \frac{\mu \cdot m \cdot g}{m} = \mu \cdot g \]
Substituting the values:
\[ a = 0.5 \cdot 10 = 5 \, \text{m/s}^2 \]
This acceleration will act in the opposite direction to the motion of the box (deceleration).

### Step 5: Use the Kinematic Equation
We can use the kinematic equation to find the distance (s) the box moves relative to the belt before coming to rest:
\[ v^2 = u^2 + 2as \]
Substituting the known values:
- \( v = 0 \)
- \( u = 2 \, \text{m/s} \)
- \( a = -5 \, \text{m/s}^2 \) (negative because it's deceleration)

The equation becomes:
\[ 0 = (2)^2 + 2(-5)s \]
\[ 0 = 4 - 10s \]

### Step 6: Solve for Distance (s)
Rearranging the equation gives:
\[ 10s = 4 \]
\[ s = \frac{4}{10} = 0.4 \, \text{m} \]

### Final Answer
The distance that the box will move relative to the belt before coming to rest is **0.4 meters**.

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