A `1.5-kg` block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of x-axis is applied to the block. The force is given by `vecF=(4-x^2)veciN`, where x is in meter and the initial position of the block is `x=0`.
The maximum kinetic energy of the block between `x=0` and `x=2.0m` is
A.2.33 J B.8.67 J C.5.33 J D.6.67 J

Options

A. 2.33 J
B. 8.67 J
C. 5.33 J
D. 6.67 J

Answer

To find the maximum kinetic energy of the block between \( x = 0 \) and \( x = 2.0 \, \text{m} \), we can use the work-energy theorem, which states that the change in kinetic energy is equal to the work done by the forces acting on the object.

### Step-by-Step Solution

1. **Identify the Force Function**:
The force acting on the block is given by:
\[ \vec{F} = (4 - x^2) \hat{i} \, \text{N} \]

2. **Work Done Calculation**:
The work done \( W \) on the block as it moves from \( x = 0 \) to \( x = 2 \) can be calculated using the integral of the force over the distance:
\[ W = \int_{0}^{2} F \, dx = \int_{0}^{2} (4 - x^2) \, dx \]

3. **Integrate the Force**:
Now we perform the integration:
\[ W = \int_{0}^{2} (4 - x^2) \, dx = \left[ 4x - \frac{x^3}{3} \right]_{0}^{2} \]

4. **Evaluate the Integral**:
Substitute the limits into the integrated function:
\[ W = \left( 4(2) - \frac{(2)^3}{3} \right) - \left( 4(0) - \frac{(0)^3}{3} \right) \]
\[ W = \left( 8 - \frac{8}{3} \right) - 0 = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3} \, \text{J} \]

5. **Kinetic Energy Calculation**:
Since the block starts from rest, the initial kinetic energy \( K_i = 0 \). The work done is equal to the change in kinetic energy:
\[ K_f = K_i + W = 0 + \frac{16}{3} \, \text{J} \]

6. **Maximum Kinetic Energy**:
The maximum kinetic energy of the block at \( x = 2 \) is:
\[ K = \frac{16}{3} \approx 5.33 \, \text{J} \]

### Conclusion
The maximum kinetic energy of the block between \( x = 0 \) and \( x = 2.0 \, \text{m} \) is approximately \( 5.33 \, \text{J} \). Thus, the correct answer is option **C**.