A particle of mass m is moving in the x y -plane such that its velocity at a point ( x ,   y ) is given…

As force is given by, $\overrightarrow{F}=m\frac{d\overrightarrow{v}}{dt}$.

Given: $\overrightarrow{v}=\alpha \left(y\stackrel{\wedge }{x}+2x\stackrel{\wedge }{y}\right)$, where velocity along x-axis is $v_x=\alpha y$ and velocity along y-axis is $v_y=2x\alpha$.

Therefore,

$\frac{d\overrightarrow{v}}{dt}=\alpha \left(\frac{dy}{dt}\stackrel{\wedge }{x}+2\frac{dx}{dt}\stackrel{\wedge }{y}\right)$

$=\alpha \left(v_y\stackrel{\wedge }{x}+2v_x\stackrel{\wedge }{y}\right)$

$=\alpha \left[2x\alpha \stackrel{\wedge }{x}+2\alpha y\stackrel{\wedge }{y}\right]$

$=2{\alpha }^2\left[x\stackrel{\wedge }{x}+y\stackrel{\wedge }{y}\right]$

Therefore, required value of $\overrightarrow{F}=2m{\alpha }^2\left[x\stackrel{\wedge }{x}+y\stackrel{\wedge }{y}\right]$.