A force `F=-k(^hati + x^hatj)` (where k is a positive constant) acts on a particle moving in the `x-y` plane. Starting from the origin, the particle is taken along the positive x-axis to the point `(a,0)` and then parallel to the y-axis to the point `(a,a)`. The total work done by the force F on the particle is
(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 \( F = -k(\hat{i} + x\hat{j}) \) on a particle moving in the \( x-y \) plane from the origin to the point \( (a, 0) \) and then to the point \( (a, a) \).

### Step 1: Define the Work Done
The work done by a force \( F \) when moving along a path is given by the integral:
\[ W = \int F \cdot dr \]
where \( dr \) is the differential displacement vector.

### Step 2: Set Up the Displacement Vector
In the \( x-y \) plane, the displacement vector can be expressed as:
\[ dr = dx \hat{i} + dy \hat{j} \]

### Step 3: Express the Force Vector
The force vector is given as:
\[ F = -k(0\hat{i} + x\hat{j}) = -k(0\hat{i} + y\hat{j}) \]

### Step 4: Calculate Work Done Along Path OA (from O to A)
1. From the origin \( O(0,0) \) to point \( A(a,0) \), the movement is along the \( x \)-axis, where \( y = 0 \).
2. Therefore, \( dy = 0 \) and \( y = 0 \) during this segment.

The work done along OA is:
\[ W_{OA} = \int_{0}^{a} F \cdot dr = \int_{0}^{a} (-k(0\hat{i} + 0\hat{j})) \cdot (dx \hat{i} + 0 \hat{j}) = \int_{0}^{a} 0 \, dx = 0 \]

### Step 5: Calculate Work Done Along Path AB (from A to B)
1. From point \( A(a,0) \) to point \( B(a,a) \), the movement is along the \( y \)-axis, where \( x = a \) is constant.
2. Therefore, \( dx = 0 \) and \( x = a \) during this segment.

The work done along AB is:
\[ W_{AB} = \int_{0}^{a} F \cdot dr = \int_{0}^{a} (-k(0\hat{i} + a\hat{j})) \cdot (0 \hat{i} + dy \hat{j}) = \int_{0}^{a} -k a \, dy = -k a \int_{0}^{a} dy = -k a^2 \]

### Step 6: Total Work Done
The total work done \( W \) is the sum of the work done along both segments:
\[ W = W_{OA} + W_{AB} = 0 + (-k a^2) = -k a^2 \]

### Conclusion
The total work done by the force \( F \) on the particle is:
\[ \boxed{-k a^2} \]