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 step by step, we need to analyze the situation involving the platform and the body placed on it.

### Step 1: Understand the Initial Conditions
- The platform is moving with a speed \( v = 6 \, \text{m/s} \).
- A body of mass \( m \) (where \( m \ll M \)) is placed on the platform at \( t = 0 \).
- 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 2: Determine the Initial Velocity of the Body
When the body is placed on the moving platform, it initially has no velocity relative to the platform. However, since the platform is moving, the body will have an initial velocity of:
\[ v_{\text{body}} = -v = -6 \, \text{m/s} \]
This negative sign indicates that the body is moving in the opposite direction to the platform's motion.

### Step 3: Calculate the Frictional Force
The frictional force acting on the body can be calculated using:
\[ F_{\text{friction}} = \mu \cdot m \cdot g \]
Substituting the values:
\[ F_{\text{friction}} = 0.3 \cdot m \cdot 10 = 3m \, \text{N} \]

### Step 4: Determine the Acceleration of the Body
Using Newton's second law, the acceleration \( a \) of the body due to friction is:
\[ F = m \cdot a \implies 3m = m \cdot a \implies a = 3 \, \text{m/s}^2 \]
This acceleration is directed opposite to the motion of the platform.

### Step 5: Use Kinematic Equation to Find Distance
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Where:
- \( v = 0 \) (final velocity when the body comes to rest),
- \( u = -6 \, \text{m/s} \) (initial velocity),
- \( a = -3 \, \text{m/s}^2 \) (acceleration, negative because it opposes the motion),
- \( s \) is the distance covered by the body.

Substituting the values into the equation:
\[ 0 = (-6)^2 + 2(-3)s \]
\[ 0 = 36 - 6s \]
\[ 6s = 36 \implies s = 6 \, \text{m} \]

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

### Final Answer
The distance covered by the body before coming to rest is \( 6 \, \text{m} \) in the direction opposite to the motion of the platform.

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