t = 2h g = 2× 1.25 10 = 0.5s

The speed when the block leaves the chute: v = RT = 2m/s;

The acceleration of the animal block is determined by Newton's second law: F-? mg=ma 1

Substituting the data, the result is: a1= f-μ GMM = 4-0.2× 0.4×100.4 = 8m/S2.

After removing the external force, Newton's second law:-μ g = Ma 2;

De: a 2 =-0.2× 10=-2m/s 2。

Every time the CD rotates, it falls into it, and the pulling time is the shortest, including:

T= 2π ω = 2π 2π = 1s，

The pulley accelerates first and then decelerates on the slide, with the following characteristics: v0 = a1t1+a2t2 = 8t1-2t2.

Therefore, 4t1-t2 =1;

The relationship between taxiing time, throwing time in the air and disk period: t 1 +t 2 +t=T,

Simultaneous and alternative data: t1= 0.3s. 。

(2) If the pulling time is 0.5s, the accelerated displacement is: x1=12a1t2 =12x8× 0.25 =1m;

The final acceleration speed is v1= a1t = 8× 0.5 = 4m/s;

The displacement during deceleration is x2 = v2-v212a2 = 4-16-4m/s = 3m;

The board length is: x = x1+x2 =1+3 = 4m;

(3) The pulling time is 0.5s, and the deceleration time is t3 = v-v1a2 = 2-4-2s =1s;

Time: T4 = 0.5+T3 = 0.5+1=1.5s;

So the number of turns n =1.51=1.5;

Answer: (1) The shortest pulling time is 0.3s；;

(2) The length of the plate is 4m;

(3) The disc rotates 65438 0.5 feet.