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

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the applied force at \( t = 2 \, s \)

The applied force \( F \) is given by the equation:
\[ F = kt^2 \]
where \( k = 2 \, \text{N/s}^2 \) and \( t = 2 \, s \).

Substituting the values:
\[ F = 2 \times (2^2) = 2 \times 4 = 8 \, \text{N} \]

### Step 2: Calculate the normal force acting on the block

The normal force \( N \) acting on the block is equal to the weight of the block, which can be calculated using:
\[ N = mg \]
where \( m = 4 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \).

Substituting the values:
\[ N = 4 \times 10 = 40 \, \text{N} \]

### Step 3: Calculate the maximum (limiting) frictional force

The maximum frictional force \( f_{\text{max}} \) can be calculated using the formula:
\[ f_{\text{max}} = \mu N \]
where \( \mu = 0.8 \).

Substituting the values:
\[ f_{\text{max}} = 0.8 \times 40 = 32 \, \text{N} \]

### Step 4: Compare the applied force with the maximum frictional force

Now we compare the applied force \( F = 8 \, \text{N} \) with the maximum frictional force \( f_{\text{max}} = 32 \, \text{N} \).

Since \( F < f_{\text{max}} \), the block will not slide, and the frictional force will equal the applied force.

### Step 5: Determine the frictional force acting on the block

Since the block is in equilibrium and the applied force is less than the maximum frictional force, the frictional force \( f \) will be equal to the applied force:
\[ f = F = 8 \, \text{N} \]

### Final Answer

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

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