A particle is moved along a path AB-BC-CD-DE-EF-FA, as shown in figure, in presence of a forced $F = (α y i + 2α x j) N,$ where x and y are in meter and $α = -1 N m^{-1}.$ The work done on the particle by this force F will be ___ Joule.

Step 1: Define the force vector and the work integral.

The given force vector is:

$$\vec{F} = \alpha y\hat{i} + 2\alpha x\hat{j}$$

Substituting $\alpha = -1\,\text{N}\,\text{m}^{-1}$ yields:

$$\vec{F} = -y\hat{i} - 2x\hat{j}$$

The total work done along the closed path is:

$$W = \oint \vec{F} \cdot d\vec{r}$$ $$W = \oint (-y\,dx - 2x\,dy)$$

Step 2: Calculate the work done along each segment.

The closed path consists of six segments: $AB$, $BC$, $CD$, $DE$, $EF$, and $FA$.

For segment $AB$ from $(0, 1)$ to $(1, 1)$, $y = 1$ and $dy = 0$:

$$W_{AB} = \int_{0}^{1} -1\,dx = -1\,\text{J}$$

For segment $BC$ from $(1, 1)$ to $(1, 0.5)$, $x = 1$ and $dx = 0$:

$$W_{BC} = \int_{1}^{0.5} -2(1)\,dy = 1\,\text{J}$$

For segment $CD$ from $(1, 0.5)$ to $(0.5, 0.5)$, $y = 0.5$ and $dy = 0$:

$$W_{CD} = \int_{1}^{0.5} -0.5\,dx = 0.25\,\text{J}$$

For segment $DE$ from $(0.5, 0.5)$ to $(0.5, 0)$, $x = 0.5$ and $dx = 0$:

$$W_{DE} = \int_{0.5}^{0} -2(0.5)\,dy = 0.5\,\text{J}$$

For segment $EF$ from $(0.5, 0)$ to $(0, 0)$, $y = 0$ and $dy = 0$:

$$W_{EF} = \int_{0.5}^{0} 0\,dx = 0\,\text{J}$$

For segment $FA$ from $(0, 0)$ to $(0, 1)$, $x = 0$ and $dx = 0$:

$$W_{FA} = \int_{0}^{1} -2(0)\,dy = 0\,\text{J}$$

Step 3: Sum the work done over all segments.

$$W = W_{AB} + W_{BC} + W_{CD} + W_{DE} + W_{EF} + W_{FA}$$ $$W = -1 + 1 + 0.25 + 0.5 + 0 + 0$$ $$W = 0.75\,\text{J}$$

Step 4: Verify using Green's Theorem.

The curl of the force field is:

$$\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = 2\alpha - \alpha = \alpha$$

Since the path is traversed clockwise, the work done is:

$$W = -\iint \alpha\,dA = -\alpha A$$

The enclosed area $A$ is:

$$A = (1 \times 1) - (0.5 \times 0.5) = 0.75\,\text{m}^2$$

Substituting $\alpha = -1\,\text{N}\,\text{m}^{-1}$ and $A = 0.75\,\text{m}^2$:

$$W = -(-1)(0.75) = 0.75\,\text{J}$$

Answer: The work done on the particle by this force is 0.75 Joule.