A varable force, given by the 2- dimensional vector`overlineF=(3xx^(2)hati+4 hatj),` acts on a particle. The force is in newton and x is in metre. What is the change in the kinetic energy of the particle as it moves from the point with coordinates (2,3) to (3,0) (The coornates are in metres)

Options

A. `-7 J`
B. zero
C. `+7 J`
D. `+19 J`

Explanation

To find the change in kinetic energy of a particle moving under the influence of a variable force, we can use the work-energy theorem, which states that the work done on the particle is equal to the change in its kinetic energy.

Given the force vector:
\[ \overline{F} = (3x^2 \hat{i} + 4 \hat{j}) \, \text{N} \]
and the particle moves from the point (2, 3) to (3, 0).

### Step 1: Set up the work done integral
The work done \( W \) by the force when the particle moves from point A to point B can be calculated using the line integral:
\[ W = \int_A^B \overline{F} \cdot d\overline{r} \]
where \( d\overline{r} = dx \hat{i} + dy \hat{j} \).

### Step 2: Substitute the force and displacement
Substituting the expression for \( \overline{F} \) and \( d\overline{r} \):
\[ W = \int_A^B (3x^2 \hat{i} + 4 \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) \]
This simplifies to:
\[ W = \int_A^B (3x^2 dx + 4 dy) \]

### Step 3: Determine the limits of integration
The particle moves from point (2, 3) to (3, 0). Therefore, we can break the integral into two parts:
1. From \( x = 2 \) to \( x = 3 \) while \( y \) changes from \( 3 \) to \( 0 \).
2. For the \( y \) component, we can express \( y \) as a function of \( x \) if necessary, but here we can integrate \( dy \) directly.

### Step 4: Calculate the work done
We can calculate the work done in two parts:
1. **For \( dx \)**:
\[ W_x = \int_{2}^{3} 3x^2 \, dx \]
\[ W_x = 3 \left[ \frac{x^3}{3} \right]_{2}^{3} = [3^3 - 2^3] = 27 - 8 = 19 \, \text{J} \]

2. **For \( dy \)**:
\[ W_y = \int_{3}^{0} 4 \, dy = 4[y]_{3}^{0} = 4(0 - 3) = -12 \, \text{J} \]

### Step 5: Combine the work done
Now, combine the two components of work done:
\[ W = W_x + W_y = 19 - 12 = 7 \, \text{J} \]

### Step 6: Apply the work-energy theorem
According to the work-energy theorem:
\[ \Delta KE = W \]
Thus, the change in kinetic energy of the particle is:
\[ \Delta KE = 7 \, \text{J} \]

### Final Answer
The change in kinetic energy of the particle as it moves from the point (2, 3) to (3, 0) is **7 Joules**.

---