A force `F = -K(y hatI + x hatj)`(where `K` is a posive constant ) acts on a particle moving in the `xy` plane . Starting form the original , the partical is taken along in the positive `x` axis to the point `(x,0)` and then partical to they axis the point `(x,0)` . The total work done by the force `F` on the particls is

Options

A. `-2Ka^(2)`
B. `2Ka^(2)`
C. `-Ka^(2)`
D. `Ka^(2)`

Explanation

To solve the problem, we need to calculate the total work done by the force \( F = -K(y \hat{i} + x \hat{j}) \) on a particle moving in the xy-plane. The particle moves from the origin to the point \( (a, 0) \) along the x-axis and then from \( (a, 0) \) to \( (a, a) \) along the y-axis.

### Step-by-Step Solution:

1. **Identify the Path of Motion**:
The particle moves in two segments:
- From \( (0, 0) \) to \( (a, 0) \) along the x-axis.
- From \( (a, 0) \) to \( (a, a) \) along the y-axis.

2. **Calculate Work Done in the First Segment (from \( (0, 0) \) to \( (a, 0) \))**:
- In this segment, the displacement \( d\mathbf{x} \) is along the x-axis, so \( d\mathbf{x} = dx \hat{i} \).
- The force \( F \) at this point is \( F = -K(0 \hat{i} + x \hat{j}) = -K(0 \hat{i} + a \hat{j}) = -K(0 \hat{i}) \) since \( y = 0 \).
- The work done \( W_1 \) is given by:
\[ W_1 = \int_{0}^{a} F \cdot d\mathbf{x} = \int_{0}^{a} (-K(0 \hat{i})) \cdot (dx \hat{i}) = 0 \]
- Since the force has no component in the direction of displacement, \( W_1 = 0 \).

3. **Calculate Work Done in the Second Segment (from \( (a, 0) \) to \( (a, a) \))**:
- In this segment, the displacement \( d\mathbf{x} \) is along the y-axis, so \( d\mathbf{x} = dy \hat{j} \).
- The force \( F \) at this point is \( F = -K(y \hat{i} + a \hat{j}) \).
- The work done \( W_2 \) is given by:
\[ W_2 = \int_{0}^{a} F \cdot d\mathbf{x} = \int_{0}^{a} (-K(y \hat{i} + a \hat{j})) \cdot (dy \hat{j}) = \int_{0}^{a} (-K a) dy \]
- Evaluating the integral:
\[ W_2 = -K a \int_{0}^{a} dy = -K a [y]_{0}^{a} = -K a^2 \]

4. **Total Work Done**:
- The total work done \( W \) is the sum of the work done in both segments:
\[ W = W_1 + W_2 = 0 - K a^2 = -K a^2 \]

### Final Answer:
The total work done by the force \( F \) on the particle is \( -K a^2 \).