A force $F = -K(y i + x j)$ (where $K$ is a positive constant) acts on a particle moving in the $xy$ 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

Step 1: Define the force and the work integral.

The force acting on the particle is given by:

$$\vec{F} = -K(y\hat{i} + x\hat{j})$$

The total work done along the path is:

$$W = \int \vec{F} \cdot d\vec{r}$$

where the displacement vector is:

$$d\vec{r} = dx\hat{i} + dy\hat{j}$$

Step 2: Calculate the work done along the first segment.

For the path from $(0,0)$ to $(a,0)$ along the $x$-axis, we have $y = 0$ and $dy = 0$.

$$d\vec{r} = dx\hat{i}$$ $$\vec{F} = -Kx\hat{j}$$ $$\vec{F} \cdot d\vec{r} = 0$$ $$W_1 = \int_{0}^{a} 0 \, dx = 0$$

Step 3: Calculate the work done along the second segment.

For the path from $(a,0)$ to $(a,a)$ parallel to the $y$-axis, we have $x = a$ and $dx = 0$.

$$d\vec{r} = dy\hat{j}$$ $$\vec{F} = -K(y\hat{i} + a\hat{j})$$ $$\vec{F} \cdot d\vec{r} = -Ka \, dy$$ $$W_2 = \int_{0}^{a} -Ka \, dy$$ $$W_2 = -Ka^2$$

Step 4: Calculate the total work done.

Sum the work done over both segments.

$$W = W_1 + W_2$$ $$W = 0 - Ka^2$$ $$W = -Ka^2$$

Step 5: Verify using the conservative nature of the force.

The curl of the force field is zero, which means the force is conservative.

$$\vec{\nabla} \times \vec{F} = 0$$

The potential energy function $U(x,y)$ associated with this force is:

$$U(x,y) = Kxy$$

The work done is the negative change in potential energy.

$$W = -\Delta U = -(U(a,a) - U(0,0))$$ $$W = -(Ka^2 - 0) = -Ka^2$$

Answer: The total work done by the force is $-Ka^2$.