Slide, Hook, Roll

The ball has mass $m$ and radius $r$ , and we know that for a solid sphere $I = \tfrac25 m r^2$ .

The ball is initially sliding with speed $v_0$ . While sliding, the kinetic friction force on the ball is $F = -\mu m g.$ On one hand, this causes the ball to decelerate translationally; on the other hand, it accelerates rotationally. $a = \frac F m = -\mu g;\ \ \ \ \ \alpha = \frac \tau I = \frac{\mu m g r}{\tfrac25 m r^2} = \frac{5\mu g}{2 r}.$ After some time $t$ , the speed of the ball is $v = v_0 - at$ and its rotational speed, $\omega = \alpha t$ . The rolling stage begins when $v = \omega r$ . This gives us the equation $v_0 - \mu g t = \frac{5\mu g}{2 r} r t\ \ \therefore\ \ t = \frac{2 v_0}{7 \mu g}.$ Substituting this, we find that the rolling speed of the ball will be $v = v_0 - \mu g\frac{2 v_0}{7 \mu g} = \tfrac 57 v_0;$ of course, $\omega r$ will have the same value. The kinetic energy at that time is $K = \tfrac12 m v^2 + \tfrac 12 I \omega^2 = \tfrac12 m (1 + \tfrac 25) (\tfrac 57v_0)^2 = \tfrac 57\cdot \tfrac12 m v_0^2.$ Compare this to the initial kinetic energy $K = \tfrac12 m v_0^2$ , and we see that the fraction of kinetic energy lost is $1 - (\tfrac 57) = \tfrac 27 \approx \boxed{29\%}.$

Note that the friction force not only converted $29\%$ of the initial translational KE into heat, but also converts $20\%$ of it into rotational KE. The remaining $51\%$ remains translational KE.