A force acting on particle is given by `vecF=(3x^2hati+4yhatj)N`. The change in kinetic energy of particle as it moves from `(0,2m)` to `(1m,3m)` is

Options

A. `6J`
B. `10J`
C. `11J`
D. `13J`

Explanation

To solve the problem of finding the change in kinetic energy of a particle moving under the influence of a variable force, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Force and Displacement**:
The force acting on the particle is given by:
\[ \vec{F} = (3x^2 \hat{i} + 4y \hat{j}) \, \text{N} \]
The particle moves from the point \( (0, 2 \, \text{m}) \) to \( (1 \, \text{m}, 3 \, \text{m}) \).

2. **Work-Energy Theorem**:
According to the work-energy theorem, the work done by the force on the particle is equal to the change in kinetic energy:
\[ W = \Delta KE \]

3. **Expression for Work Done**:
The work done by a force during a displacement can be expressed as:
\[ W = \int \vec{F} \cdot d\vec{r} \]
where \( d\vec{r} \) is the differential displacement vector.

4. **Parameterize the Path**:
The displacement \( d\vec{r} \) can be expressed in terms of \( dx \) and \( dy \):
\[ d\vec{r} = dx \hat{i} + dy \hat{j} \]

5. **Dot Product of Force and Displacement**:
Now, we can calculate the dot product:
\[ \vec{F} \cdot d\vec{r} = (3x^2 \hat{i} + 4y \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) = 3x^2 \, dx + 4y \, dy \]

6. **Set Up the Integral**:
The work done can now be calculated by integrating from the initial to the final position:
\[ W = \int_{(0, 2)}^{(1, 3)} (3x^2 \, dx + 4y \, dy) \]
We will do this in two parts: one for \( x \) from 0 to 1 and the other for \( y \) from 2 to 3.

7. **Calculate the Work Done**:
We can split the integral:
\[ W = \int_{0}^{1} 3x^2 \, dx + \int_{2}^{3} 4y \, dy \]

- **First Integral**:
\[ \int_{0}^{1} 3x^2 \, dx = 3 \left[ \frac{x^3}{3} \right]_{0}^{1} = 3 \left( \frac{1^3}{3} - 0 \right) = 1 \]

- **Second Integral**:
\[ \int_{2}^{3} 4y \, dy = 4 \left[ \frac{y^2}{2} \right]_{2}^{3} = 4 \left( \frac{3^2}{2} - \frac{2^2}{2} \right) = 4 \left( \frac{9}{2} - \frac{4}{2} \right) = 4 \left( \frac{5}{2} \right) = 10 \]

8. **Total Work Done**:
Now, add the results of the two integrals:
\[ W = 1 + 10 = 11 \, \text{J} \]

9. **Change in Kinetic Energy**:
From the work-energy theorem:
\[ \Delta KE = W = 11 \, \text{J} \]

### Final Answer:
The change in kinetic energy of the particle as it moves from \( (0, 2 \, \text{m}) \) to \( (1 \, \text{m}, 3 \, \text{m}) \) is:
\[ \Delta KE = 11 \, \text{J} \]