A force `F=-K(yhati+xhatj)` (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

Options

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

Answer

To find the total work done by the force \( F = -K(y \hat{i} + x \hat{j}) \) on a particle moving from the origin to the point \( (a, 0) \) and then to \( (a, a) \), we can follow these steps:

### Step 1: Define the Force
The force acting on the particle is given as:
\[ F = -K(y \hat{i} + x \hat{j}) \]
where \( K \) is a positive constant.

### Step 2: Break the Path into Two Segments
The particle moves in two segments:
1. From the origin \( (0, 0) \) to \( (a, 0) \) along the x-axis.
2. From \( (a, 0) \) to \( (a, a) \) along the y-axis.

### Step 3: Calculate Work Done in the First Segment
For the first segment, the particle moves along the x-axis where \( y = 0 \). Thus, the force simplifies to:
\[ F = -K(0 \hat{i} + x \hat{j}) = -Kx \hat{j} \]
The displacement vector \( d\mathbf{r} \) during this segment is:
\[ d\mathbf{r} = dx \hat{i} \]
The work done \( dW \) in this segment is:
\[ dW = F \cdot d\mathbf{r} = (-Kx \hat{j}) \cdot (dx \hat{i}) = 0 \]
Since the force is perpendicular to the displacement, the work done in this segment is:
\[ W_1 = 0 \]

### Step 4: Calculate Work Done in the Second Segment
For the second segment, the particle moves along the y-axis where \( x = a \). Thus, the force simplifies to:
\[ F = -K(y \hat{i} + a \hat{j}) = -Ky \hat{i} - Ka \hat{j} \]
The displacement vector \( d\mathbf{r} \) during this segment is:
\[ d\mathbf{r} = dy \hat{j} \]
The work done \( dW \) in this segment is:
\[ dW = F \cdot d\mathbf{r} = (-Ka \hat{j}) \cdot (dy \hat{j}) = -Ka \, dy \]
To find the total work done in this segment, we integrate from \( y = 0 \) to \( y = a \):
\[ W_2 = \int_0^a -Ka \, dy = -Ka \cdot y \bigg|_0^a = -Ka^2 \]

### Step 5: Total Work Done
The total work done \( W \) is the sum of the work done in both segments:
\[ W = W_1 + W_2 = 0 - Ka^2 = -Ka^2 \]

### Final Answer
Thus, the total work done by the force \( F \) on the particle is:
\[ \boxed{-Ka^2} \]