A block of mass m is gently placed over a massive plank moving horizontal over a smooth surface with velocity `10 ms^(-1)`. The coefficient of friction between the block and the plank is 0.2. The distance travelled by the block till it slides on the plank is `[g = 10 ms^(-2)]`

Options

A. 10 m
B. 15m
C. 25 m
D. 35m

Answer

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the Problem
We have a block of mass \( m \) placed on a plank that is moving horizontally with a velocity of \( 10 \, \text{m/s} \). The coefficient of friction between the block and the plank is \( \mu = 0.2 \). We need to find the distance traveled by the block until it starts sliding on the plank.

### Step 2: Analyze the Forces
When the block is placed on the moving plank, it initially has a velocity of \( 0 \, \text{m/s} \) relative to the plank. The frictional force acting on the block is what allows it to accelerate and catch up to the velocity of the plank.

The frictional force \( F_f \) can be given by:
\[ F_f = \mu m g \]
where \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).

### Step 3: Calculate the Acceleration
The acceleration \( a \) of the block due to the frictional force is:
\[ a = \frac{F_f}{m} = \frac{\mu m g}{m} = \mu g \]
Substituting the values:
\[ a = 0.2 \times 10 = 2 \, \text{m/s}^2 \]

### Step 4: Use Kinematic Equations
We can use the kinematic equation to find the distance \( s \) traveled by the block until it reaches the velocity of the plank:
\[ v^2 = u^2 + 2as \]
Where:
- \( v = 10 \, \text{m/s} \) (final velocity of the block when it matches the plank's speed)
- \( u = 0 \, \text{m/s} \) (initial velocity of the block)
- \( a = 2 \, \text{m/s}^2 \) (acceleration)

Substituting the values into the equation:
\[ (10)^2 = (0)^2 + 2 \times 2 \times s \]
\[ 100 = 4s \]
\[ s = \frac{100}{4} = 25 \, \text{m} \]

### Step 5: Conclusion
The distance traveled by the block until it starts sliding on the plank is \( 25 \, \text{m} \).