A horizontal force just sufficient to move a body of mass `4 kg ` lying on a rought horizontal surface is applied on it .The coefficient of static and kinetic friction the body and the surface are `0.8 and 0.6` respectively If the force contines to act even after the block has started moving the acceleration of the block in `ms^(-2)` is `(g = 10 ms^(-2))`

Options

A. `1//4`
B. `1//2`
C. `2`
D. `4`

Explanation

To solve the problem, we need to determine the acceleration of the block after it starts moving. We will follow these steps:

### Step 1: Calculate the Normal Force (N)
The block is lying on a horizontal surface, so the normal force (N) is equal to the weight of the block.

\[ N = mg \]

Given:
- Mass \( m = 4 \, \text{kg} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

Calculating the normal force:

\[ N = 4 \, \text{kg} \times 10 \, \text{m/s}^2 = 40 \, \text{N} \]

### Step 2: Calculate the Limiting Friction Force (F_L)
The limiting friction force is given by:

\[ F_L = \mu_s \times N \]

Where:
- Coefficient of static friction \( \mu_s = 0.8 \)

Calculating the limiting friction force:

\[ F_L = 0.8 \times 40 \, \text{N} = 32 \, \text{N} \]

### Step 3: Calculate the Kinetic Friction Force (F_K)
Once the block starts moving, the kinetic friction force acts on it. This is given by:

\[ F_K = \mu_k \times N \]

Where:
- Coefficient of kinetic friction \( \mu_k = 0.6 \)

Calculating the kinetic friction force:

\[ F_K = 0.6 \times 40 \, \text{N} = 24 \, \text{N} \]

### Step 4: Apply Newton's Second Law
When the block is moving, the net force acting on it can be expressed as:

\[ F_{\text{net}} = F - F_K \]

Where \( F \) is the applied force (which is equal to the limiting friction force when it starts moving, hence \( F = 32 \, \text{N} \)).

Substituting the values:

\[ F_{\text{net}} = 32 \, \text{N} - 24 \, \text{N} = 8 \, \text{N} \]

### Step 5: Calculate the Acceleration (a)
Using Newton's second law \( F = ma \):

\[ F_{\text{net}} = ma \]

Substituting the values:

\[ 8 \, \text{N} = 4 \, \text{kg} \times a \]

Solving for \( a \):

\[ a = \frac{8 \, \text{N}}{4 \, \text{kg}} = 2 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the block is \( 2 \, \text{m/s}^2 \).

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