A block of mass `4kg` is placed on a rough horizontal plane A time dependent force `F = kt^(2)` acts on the block where `k = 2N//s^(2)` Coefficient of friction `mu = 0.8` force of friction between the block and the plane at `t = 2s` is

Options

A. `32N`
B. `4N`
C. `2N`
D. `8N`

Explanation

To solve the problem step by step, we will calculate the force of friction acting on the block at \( t = 2 \) seconds.

### Step 1: Calculate the Normal Force
The normal force \( N \) acting on the block can be calculated using the formula:
\[ N = mg \]
where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

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 Maximum Force of Friction
The maximum force of friction \( f_{\text{max}} \) can be calculated using the formula:
\[ f_{\text{max}} = \mu N \]
where \( \mu \) is the coefficient of friction.

Given:
- Coefficient of friction \( \mu = 0.8 \)

Calculating the maximum force of friction:
\[ f_{\text{max}} = 0.8 \times 40 \, \text{N} = 32 \, \text{N} \]

### Step 3: Calculate the Applied Force at \( t = 2 \) seconds
The applied force \( F \) is given by the equation:
\[ F = kt^2 \]
where \( k \) is a constant.

Given:
- \( k = 2 \, \text{N/s}^2 \)
- \( t = 2 \, \text{s} \)

Calculating the applied force:
\[ F = 2 \, \text{N/s}^2 \times (2 \, \text{s})^2 = 2 \times 4 = 8 \, \text{N} \]

### Step 4: Determine the Force of Friction
Since the applied force \( F \) (8 N) is less than the maximum force of friction \( f_{\text{max}} \) (32 N), the force of friction will equal the applied force to prevent motion.

Thus, the force of friction \( f \) at \( t = 2 \) seconds is:
\[ f = F = 8 \, \text{N} \]

### Final Answer
The force of friction between the block and the plane at \( t = 2 \) seconds is \( 8 \, \text{N} \).

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