A massive platform of mass M is moving with speed `v=6ms^(-1)` At t = 0, at body of mass `m(m lt lt M)` is gently placed on the platform. If c-efficient of friction between the body and platform is `mu=0.3 and g=10 m//s^(2)`, then

Options

A. the body covers a distance of 3 m on the platform in the direction of motion of the platform.
B. the body covers a distance of 3 m on the platform opposite to the direction of motion of the platform before coming to rest.
C. the body covers a distance of 6 m on the platform in the direction of motion of the platform.
D. the body covers a distance of 6 m on the platform opposite to the direction of motion of the platform before coming to rest.

Explanation

To solve the problem, we need to analyze the motion of the body of mass \( m \) placed on the moving platform of mass \( M \). The platform is moving with a speed \( v = 6 \, \text{m/s} \), and the coefficient of friction between the body and the platform is \( \mu = 0.3 \). The acceleration due to gravity is \( g = 10 \, \text{m/s}^2 \).

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The platform is moving with a speed \( v = 6 \, \text{m/s} \).
- The body of mass \( m \) is placed gently on the platform, which means its initial relative velocity with respect to the platform is \( 0 \, \text{m/s} \).

2. **Determine Relative Velocity**:
- The initial velocity of the body \( m \) with respect to the ground is \( 6 \, \text{m/s} \) (same as the platform).
- When the body is placed on the platform, it starts with an initial velocity of \( 0 \) relative to the platform. Thus, the relative velocity of the body with respect to the platform is:
\[ v_{21} = v_{2} - v_{1} = 0 - 6 = -6 \, \text{m/s} \]
- This indicates that the body is moving leftward with respect to the platform.

3. **Friction Force**:
- The friction force \( F_k \) acting on the body is given by:
\[ F_k = \mu \cdot N \]
- The normal force \( N \) acting on the body is equal to its weight, which is:
\[ N = m \cdot g \]
- Therefore, the frictional force becomes:
\[ F_k = \mu \cdot m \cdot g = 0.3 \cdot m \cdot 10 = 3m \, \text{N} \]
- This frictional force acts in the rightward direction (opposite to the leftward relative motion).

4. **Calculate Acceleration**:
- According to Newton's second law, the acceleration \( a \) of the body can be calculated as:
\[ F_k = m \cdot a \implies 3m = m \cdot a \implies a = 3 \, \text{m/s}^2 \]
- The acceleration is positive, indicating that it acts in the rightward direction.

5. **Use the Equation of Motion**:
- We need to find the distance \( x \) covered by the body before it comes to rest. The final velocity \( v = 0 \) when the body stops.
- Using the equation of motion:
\[ v^2 = u^2 + 2ax \]
- Here, \( u = -6 \, \text{m/s} \) (initial velocity), \( v = 0 \), and \( a = 3 \, \text{m/s}^2 \):
\[ 0 = (-6)^2 + 2 \cdot (-3) \cdot x \]
\[ 0 = 36 - 6x \]
\[ 6x = 36 \implies x = 6 \, \text{m} \]

6. **Conclusion**:
- The body covers a distance of \( 6 \, \text{m} \) on the platform before coming to rest.

### Final Answer:
The body covers a distance of \( 6 \, \text{m} \) on the platform opposite to the direction of motion of the platform before coming to rest.

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