A suitcase is gently dropped on a conveyor belt moving at `3ms^(-1)` If the coefficient of friction between the belt and suitcase is `0.5` how far will the suitcase move on the belt before coming to rest ?

Explanation

To solve the problem of how far the suitcase will move on the conveyor belt before coming to rest, we can follow these steps:

### Step 1: Identify the Given Values
- Initial velocity of the conveyor belt, \( v = 3 \, \text{m/s} \)
- Coefficient of friction, \( \mu = 0.5 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (assuming standard value)

### Step 2: Calculate the Frictional Force
The frictional force \( F_f \) acting on the suitcase can be calculated using the formula:
\[ F_f = \mu \cdot m \cdot g \]
Where \( m \) is the mass of the suitcase. However, since we are looking for distance, we can express the acceleration caused by this frictional force without needing the mass.

### Step 3: Calculate the Acceleration
The acceleration \( a \) due to the frictional force can be calculated as:
\[ a = \frac{F_f}{m} = \frac{\mu \cdot m \cdot g}{m} = \mu \cdot g \]
Substituting the values:
\[ a = 0.5 \cdot 10 = 5 \, \text{m/s}^2 \]

### Step 4: Use Kinematic Equation
We will use the kinematic equation to find the distance \( s \) that the suitcase travels before coming to rest:
\[ v^2 = u^2 + 2as \]
Where:
- \( v = 0 \) (final velocity when the suitcase comes to rest)
- \( u = 3 \, \text{m/s} \) (initial velocity of the suitcase)
- \( a = -5 \, \text{m/s}^2 \) (negative because it is deceleration)

Substituting the values into the equation:
\[ 0 = (3)^2 + 2(-5)s \]
\[ 0 = 9 - 10s \]

### Step 5: Solve for Distance \( s \)
Rearranging the equation gives:
\[ 10s = 9 \implies s = \frac{9}{10} = 0.9 \, \text{m} \]

### Conclusion
The suitcase will move a distance of \( 0.9 \, \text{m} \) on the conveyor belt before coming to rest.

---