A solid sphere of mass $1$kg rolls on a table with linear speed $2$ m$/$s, find its total kinetic energy.

Mass of the solid sphere $m=1$ kg

Linear velocity $v=2m/s$

Total kinetic energy $K.E=K.E_{translational}+K.E_{rotational}$

$\therefore$ $K.E=\frac{1}{2}m{v}^2+\frac{1}{2}I{w}^2$

$K.E=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{r}^2\right)w^2$

As the solid sphere rolls on the table, thus $v=rw$

We get $K.E=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2$

Total kinetic energy $K.E=\frac{7}{10}m{v}^2$

$\therefore$ $K.E=\frac{7}{10}\times 1\times 2^2=2.8$ $J$