20 12 putuo mathematical model, grade two, grade three.

Friction on small blocks of wood provides centripetal force, including:

f=m4π2rT2

According to the geometric relationship, the horizontal displacement of the block after throwing x=s2? r2

The wood block falls freely in the vertical direction, t=2hg.

So the speed of the square before flying out is v = XT = S2? r22hg

Using the kinetic energy theorem in the process of wood flying out, it is concluded that:

EK- 12mv2=mgh

Solution: EK=mg(h+s2? r24h)

So, the answer is: m4π 2rT2；; mg(h+s2？ r24h)