A plank of mass `2 kg` and length `1 m` is placed on horizontal floor.A small block of mass `1 kg` is placed on top of the plank , at its right extreme end .The coefficient of friction between plank and floor is `0.5` and that between plank and block is `0.2` . If a horizontal force `= 30 N` starts acting on the plank to the right ,the time after which the block will fall off the plank is `(g = 10 ms^(-2))`

Options

A. `((2)/(3)) s`
B. `1.5 s`
C. `0.75 s`
D. `((4)/(3)) s`

Explanation

To solve the problem step by step, we will analyze the forces acting on both the plank and the block, calculate their accelerations, and then determine the time it takes for the block to fall off the plank.

### Step 1: Identify the forces acting on the block and the plank
- The plank has a mass of \( m_1 = 2 \, \text{kg} \).
- The block has a mass of \( m_2 = 1 \, \text{kg} \).
- The horizontal force acting on the plank is \( F = 30 \, \text{N} \).
- The coefficient of friction between the plank and the block is \( \mu_1 = 0.2 \).
- The coefficient of friction between the plank and the floor is \( \mu_2 = 0.5 \).
- The gravitational acceleration is \( g = 10 \, \text{m/s}^2 \).

### Step 2: Calculate the maximum frictional force between the block and the plank
The normal force \( N_1 \) acting on the block due to gravity is:
\[ N_1 = m_2 \cdot g = 1 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 10 \, \text{N} \]
The maximum frictional force \( F_{1,\text{max}} \) between the block and the plank is given by:
\[ F_{1,\text{max}} = \mu_1 \cdot N_1 = 0.2 \cdot 10 \, \text{N} = 2 \, \text{N} \]

### Step 3: Calculate the maximum frictional force between the plank and the ground
The normal force \( N_2 \) acting on the plank is:
\[ N_2 = (m_1 + m_2) \cdot g = (2 \, \text{kg} + 1 \, \text{kg}) \cdot 10 \, \text{m/s}^2 = 30 \, \text{N} \]
The maximum frictional force \( F_{2,\text{max}} \) between the plank and the ground is:
\[ F_{2,\text{max}} = \mu_2 \cdot N_2 = 0.5 \cdot 30 \, \text{N} = 15 \, \text{N} \]

### Step 4: Determine the acceleration of the plank
The net force acting on the plank is:
\[ F_{\text{net}} = F - F_{2,\text{friction}} = 30 \, \text{N} - F_{2,\text{friction}} - F_{1,\text{friction}} \]
Where \( F_{2,\text{friction}} \) is the friction force opposing the motion of the plank. The maximum friction force that can act between the block and the plank is \( 2 \, \text{N} \), so:
\[ F_{\text{net}} = 30 \, \text{N} - 15 \, \text{N} - 2 \, \text{N} = 13 \, \text{N} \]
The acceleration \( a_2 \) of the plank is:
\[ a_2 = \frac{F_{\text{net}}}{m_1} = \frac{13 \, \text{N}}{2 \, \text{kg}} = 6.5 \, \text{m/s}^2 \]

### Step 5: Determine the acceleration of the block
The block will accelerate due to the frictional force acting on it, which is the maximum frictional force \( F_{1,\text{max}} \):
\[ a_1 = \frac{F_{1,\text{max}}}{m_2} = \frac{2 \, \text{N}}{1 \, \text{kg}} = 2 \, \text{m/s}^2 \]

### Step 6: Calculate the relative acceleration
The relative acceleration \( a_{\text{relative}} \) between the plank and the block is:
\[ a_{\text{relative}} = a_2 - a_1 = 6.5 \, \text{m/s}^2 - 2 \, \text{m/s}^2 = 4.5 \, \text{m/s}^2 \]

### Step 7: Calculate the time taken for the block to fall off the plank
The distance \( s \) that the block needs to travel to fall off the plank is equal to the length of the plank, which is \( 1 \, \text{m} \). Using the equation of motion:
\[ s = u t + \frac{1}{2} a_{\text{relative}} t^2 \]
Since the initial velocity \( u = 0 \):
\[ 1 = 0 + \frac{1}{2} \cdot 4.5 \cdot t^2 \]
\[ 1 = 2.25 t^2 \]
\[ t^2 = \frac{1}{2.25} = \frac{4}{9} \]
\[ t = \frac{2}{3} \, \text{s} \]

### Final Answer
The time after which the block will fall off the plank is \( \frac{2}{3} \) seconds.