Eppur si muove

The initial energy of the ball is given by: $\displaystyle E_0 = \frac{1}{2}\ Iw_{0}^2 = \frac{1}{5}mR^2w_0^2$

When put on the surface, a frictional force $F$ acts upon it, this causes the center of the gravity to accelerate (given by $a = \frac{F}{m}$ ), and the angular to decelerate, given by $\displaystyle \varepsilon =\frac{M}{I} =\frac{FR}{\frac{2}{5}mR^2} =\frac{5}{2R}a$

The ball stops slipping, after the circumferential speed and the speed of its center of gravity are equal. Therefore it must be true that: $(w_0 - \varepsilon t)R = at$

After plugging in $\varepsilon$ , we get $\displaystyle \small t =\frac{2}{7}\frac{Rw_0}{a}$

From which we get the total speed: $\displaystyle v_f = at =\frac{2}{7}Rw_0 \hspace{6cm} \small \color{#3D99F6}w_f =\frac{v_f}{R}$

The total energy after slipping will be: $\displaystyle E_f = E_k + E_r =\frac{1}{2}mv_f^2 + \frac{1}{2}Iw_f^2 =\frac{1}{2}mR^2w_f^2 + \frac{1}{5}mR^2w_f^2 =\frac{7}{10}mR^2\frac{4}{49}w_0^2 =\frac{2}{35}mR^2w_0^2 =\frac{2}{7}E_0$

The searched energy is:

$\displaystyle k =\frac{E_0 - E_f}{E_0} =\boxed{\frac{5}{7}} \approx\boxed{0.71}$