Calculate Force Acting on Particle Moving in XY Plane

To find the force $\vec{F}$ acting on the particle, we first need to determine its acceleration $\vec{a}$.

The velocity of the particle at any point $(x, y)$ is given by: $\vec{v} = v_x \hat{x} + v_y \hat{y} = \alpha y \hat{x} + 2\alpha x \hat{y}$

Thus, the components of velocity are: $v_x = \frac{dx}{dt} = \alpha y$ $v_y = \frac{dy}{dt} = 2\alpha x$

The acceleration components are obtained by differentiating the velocity components with respect to time $t$: $a_x = \frac{dv_x}{dt} = \frac{d}{dt}(\alpha y) = \alpha \frac{dy}{dt} = \alpha (2\alpha x) = 2\alpha^2 x$ $a_y = \frac{dv_y}{dt} = \frac{d}{dt}(2\alpha x) = 2\alpha \frac{dx}{dt} = 2\alpha (\alpha y) = 2\alpha^2 y$

Therefore, the acceleration vector $\vec{a}$ is: $\vec{a} = a_x \hat{x} + a_y \hat{y} = 2\alpha^2 x \hat{x} + 2\alpha^2 y \hat{y} = 2\alpha^2 (x\hat{x} + y\hat{y})$

Using Newton's second law, the net force $\vec{F}$ acting on the particle of mass $m$ is: $\vec{F} = m\vec{a} = 2m\alpha^2 (x\hat{x} + y\hat{y})$

Hence, the correct option is (A).