A block of mass `2 kg` is initially at rest on a horizontal frictionless surface. A horizontal froce `overlineF = (9 -x^(2)) overlinei` newton acts on it, when the block is at `x = 0`. The maximum kinetic energy of the block between `x= 0 m` and `x = 3 m` in joule is
Conservation of mechanical energ

Options

A. 24
B. 20
C. 18
D. 15

Explanation

To solve the problem step by step, we will follow the work-energy principle and calculate the work done by the force acting on the block. Here’s how we can derive the maximum kinetic energy of the block:

### Step 1: Identify the Force Acting on the Block
The force acting on the block is given as:
\[ \overline{F} = (9 - x^2) \overline{i} \text{ Newton} \]
This means that the force varies with the position \(x\).

### Step 2: Determine the Work Done by the Force
To find the work done by the force as the block moves from \(x = 0\) to \(x = 3\), we need to calculate the integral of the force with respect to \(x\):
\[ W = \int_{0}^{3} F \, dx = \int_{0}^{3} (9 - x^2) \, dx \]

### Step 3: Perform the Integration
Now, we will perform the integration:
\[ W = \int_{0}^{3} (9 - x^2) \, dx = \left[ 9x - \frac{x^3}{3} \right]_{0}^{3} \]
Calculating the limits:
\[ W = \left( 9(3) - \frac{(3)^3}{3} \right) - \left( 9(0) - \frac{(0)^3}{3} \right) \]
\[ W = (27 - 9) - 0 = 18 \text{ Joules} \]

### Step 4: Apply the Work-Energy Theorem
According to the work-energy theorem, the work done on the block is equal to the change in kinetic energy:
\[ W = K_f - K_i \]
Since the block starts from rest, the initial kinetic energy \(K_i = 0\). Therefore:
\[ W = K_f - 0 \implies K_f = W \]
Thus, the final kinetic energy \(K_f\) is:
\[ K_f = 18 \text{ Joules} \]

### Conclusion
The maximum kinetic energy of the block between \(x = 0\) m and \(x = 3\) m is:
\[ \boxed{18 \text{ Joules}} \]