1. Understand the Problem and Formulate the Approach

The work done by a variable force $\vec{F}$ on a particle moving along a path is given by the line integral: $W = \int \vec{F} \cdot d\vec{r}$ where $d\vec{r} = dx \,\widehat{i} + dy \,\widehat{j}$.

The given force is $\vec{F} = (\alpha y\widehat i + 2\alpha x\widehat j)$ N. We are given $\alpha = -1$ Nm⁻¹, so the force becomes: $\vec{F} = (-y\widehat i - 2x\widehat j) \text{ N}$ The dot product $\vec{F} \cdot d\vec{r}$ is: $\vec{F} \cdot d\vec{r} = (-y\widehat i - 2x\widehat j) \cdot (dx \,\widehat{i} + dy \,\widehat{j}) = -y\,dx - 2x\,dy$ So, the work done is $W = \int (-y\,dx - 2x\,dy)$.

The total work done along the closed path AB-BC-CD-DE-EF-FA is the sum of the work done along each segment: $W_{total} = W_{AB} + W_{BC} + W_{CD} + W_{DE} + W_{EF} + W_{FA}$

The coordinates of the vertices are: A = (0, 0), B = (0, 1), C = (1, 1), D = (1, 0.5), E = (0.5, 0.5), F = (0.5, 0).

2. Calculate Work Done for Each Segment

We will calculate the work done for each straight-line segment of the path.

a) Path AB: from A(0, 0) to B(0, 1) Along this path, $x = 0$, which means $dx = 0$. The variable $y$ changes from 0 to 1. $W_{AB} = \int_A^B (-y\,dx - 2x\,dy) = \int_{y=0}^{y=1} (-y(0) - 2(0)dy) = \int_0^1 0 \,dy = 0 \text{ J}$

b) Path BC: from B(0, 1) to C(1, 1) Along this path, $y = 1$, which means $dy = 0$. The variable $x$ changes from 0 to 1. $W_{BC} = \int_B^C (-y\,dx - 2x\,dy) = \int_{x=0}^{x=1} (-(1)dx - 2x(0)) = \int_0^1 -1 \,dx = [-x]_0^1 = -(1-0) = -1 \text{ J}$

c) Path CD: from C(1, 1) to D(1, 0.5) Along this path, $x = 1$, which means $dx = 0$. The variable $y$ changes from 1 to 0.5. $W_{CD} = \int_C^D (-y\,dx - 2x\,dy) = \int_{y=1}^{y=0.5} (-y(0) - 2(1)dy) = \int_1^{0.5} -2 \,dy = [-2y]_1^{0.5} = -2(0.5) - (-2)(1) = -1 + 2 = 1 \text{ J}$

d) Path DE: from D(1, 0.5) to E(0.5, 0.5) Along this path, $y = 0.5$, which means $dy = 0$. The variable $x$ changes from 1 to 0.5. $W_{DE} = \int_D^E (-y\,dx - 2x\,dy) = \int_{x=1}^{x=0.5} (-(0.5)dx - 2x(0)) = \int_1^{0.5} -0.5 \,dx = [-0.5x]_1^{0.5} = -0.5(0.5) - (-0.5)(1) = -0.25 + 0.5 = 0.25 \text{ J}$

e) Path EF: from E(0.5, 0.5) to F(0.5, 0) Along this path, $x = 0.5$, which means $dx = 0$. The variable $y$ changes from 0.5 to 0. $W_{EF} = \int_E^F (-y\,dx - 2x\,dy) = \int_{y=0.5}^{y=0} (-y(0) - 2(0.5)dy) = \int_{0.5}^0 -1 \,dy = [-y]_{0.5}^0 = -(0) - (-0.5) = 0.5 \text{ J}$

f) Path FA: from F(0.5, 0) to A(0, 0) Along this path, $y = 0$, which means $dy = 0$. The variable $x$ changes from 0.5 to 0. $W_{FA} = \int_F^A (-y\,dx - 2x\,dy) = \int_{x=0.5}^{x=0} (-(0)dx - 2x(0)) = \int_{0.5}^0 0 \,dx = 0 \text{ J}$

3. Calculate Total Work Done

The total work done is the sum of the work done on each segment: $W_{total} = W_{AB} + W_{BC} + W_{CD} + W_{DE} + W_{EF} + W_{FA}$ $W_{total} = 0 + (-1) + 1 + 0.25 + 0.5 + 0$ $W_{total} = 0.75 \text{ J}$

Thus, the work done on the particle by the force F is 0.75 Joule.