A block of mass `2 kg` si kept at rest at `x = 0`, on a rough horizontal surface with coefficient of a rough horizontal surface with coefficient of friction `0.2`. A position dependent horizontal force given by `F = x^(2)+2x+5` (where x is in meter and `F` is in Newton), starts acting on the block at `x = 0`. The kenetic energy of block at `x = 3` is :

Options

A. `21 J`
B. `33J`
C. `45 J`
D. `12J`

Explanation

To find the kinetic energy of the block at \( x = 3 \), we will follow these steps:

### Step 1: Determine the Forces Acting on the Block
The block experiences a position-dependent force given by:
\[ F(x) = x^2 + 2x + 5 \]
The normal force \( N \) acting on the block is equal to its weight since it is on a horizontal surface:
\[ N = mg = 2 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} \]
The limiting frictional force \( f_{\text{lim}} \) can be calculated using the coefficient of friction \( \mu = 0.2 \):
\[ f_{\text{lim}} = \mu N = 0.2 \times 19.6 \, \text{N} = 3.92 \, \text{N} \]

### Step 2: Analyze the Force at \( x = 3 \)
We need to check if the applied force \( F(x) \) exceeds the limiting friction. Let's calculate \( F(3) \):
\[ F(3) = 3^2 + 2 \cdot 3 + 5 = 9 + 6 + 5 = 20 \, \text{N} \]
Since \( 20 \, \text{N} > 3.92 \, \text{N} \), the block will move, and the frictional force will act in the opposite direction to the applied force.

### Step 3: Calculate the Net Force
The net force \( F_{\text{net}} \) acting on the block can be calculated as:
\[ F_{\text{net}} = F(x) - f_{\text{lim}} = 20 \, \text{N} - 3.92 \, \text{N} = 16.08 \, \text{N} \]

### Step 4: Relate Force to Acceleration
Using Newton's second law, we can relate the net force to the acceleration \( a \):
\[ F_{\text{net}} = ma \implies a = \frac{F_{\text{net}}}{m} = \frac{16.08 \, \text{N}}{2 \, \text{kg}} = 8.04 \, \text{m/s}^2 \]

### Step 5: Use Work-Energy Principle
The work done on the block will equal the change in kinetic energy. The work done \( W \) can be calculated as:
\[ W = F_{\text{net}} \cdot d = 16.08 \, \text{N} \cdot 3 \, \text{m} = 48.24 \, \text{J} \]
Since the block starts from rest, the initial kinetic energy \( KE_i = 0 \). Thus, the final kinetic energy \( KE_f \) at \( x = 3 \) is:
\[ KE_f = KE_i + W = 0 + 48.24 \, \text{J} = 48.24 \, \text{J} \]

### Final Answer
The kinetic energy of the block at \( x = 3 \) is:
\[ \boxed{48.24 \, \text{J}} \]