A solid sphere of mass 2   kg is making pure rolling on a horizontal surface with kinetic energy 2240…

Let velocity of centre of mass of sphere be $v$.

Total kinetic energy of solid sphere is $KE=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$, where, $I$ is moment of inertia of the solid sphere about the axis passing through it's centre of mass and $\omega =\frac{v}{R} \left(\text{pure rolling condition}\right)$.

Therefore,

 $2240=\frac{1}{2}2(v{)}^2+\frac{1}{2}\left[\frac{2}{5}\left(2\right)R^2\right]{\left(\frac{v}{R}\right)}^2$

$\Rightarrow 2240=v^2+\frac{2}{5}{v}^2$

$\Rightarrow v=40 \mathrm{m} {\mathrm{s}}^{-1}$.