A suitcase is gently dropped on a conveyor belt moving at ` 3 m//s`. If the coefficient of friction between the belt and the suitcase is 0.5. Find the displacement of suitcase relative to conveyor belt before the slipping between the two is stopped `(g=10 m//s^(2))`

Options

A. 2.7 m
B. 1.8 m
C. 0.9 m
D. 1.2 m

Answer

To solve the problem step by step, we will analyze the motion of the suitcase on the conveyor belt and apply the principles of physics, particularly the concepts of friction and kinematics.

### Step 1: Identify the Given Data
- Initial velocity of the suitcase, \( u = 3 \, \text{m/s} \) (the same as the conveyor belt)
- Coefficient of friction, \( \mu = 0.5 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the Maximum Static Friction
The maximum static friction force can be calculated using the formula:
\[ f_{\text{max}} = \mu \cdot m \cdot g \]
However, we will need the acceleration due to friction to find the displacement. The frictional force will cause a deceleration (retardation) of the suitcase.

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

### Step 4: Use Kinematic Equation to Find Displacement
We will use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:
\[ v^2 = u^2 + 2as \]
Where:
- \( v = 0 \, \text{m/s} \) (final velocity when slipping stops)
- \( u = 3 \, \text{m/s} \) (initial velocity)
- \( a = -5 \, \text{m/s}^2 \) (acceleration)

Rearranging the equation to solve for displacement \( s \):
\[ 0 = (3)^2 + 2(-5)s \]
\[ 0 = 9 - 10s \]
\[ 10s = 9 \]
\[ s = \frac{9}{10} = 0.9 \, \text{m} \]

### Step 5: Conclusion
The displacement of the suitcase relative to the conveyor belt before slipping stops is \( 0.9 \, \text{m} \).

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