A, in the process of falling, at first, due to MGS+ micron ·GCOS 30 = 54mg > mg in 30, the resultant force acting on the wood block is upward, and the object slows down. In the process of falling, the resultant force is getting smaller and smaller. When the acceleration is equal to zero, the speed is minimum, the resultant force is downward, the acceleration is downward, and the speed increases, so the speed of the object decreases first and then increases. So, A is wrong.

B, when the acceleration is equal to zero, the speed is minimum, and the sliding distance of the soft rope is x, then:

l？ xLmgsin30 +L？ xLμmgcos30 =mg+xLmg

Bring in data: x= 19L, so b is correct;

C, when the block is not put, the height from the center of gravity of the soft rope to the top of the inclined plane is H 1 = L2SIN30 = L4, while when the soft rope just leaves the inclined plane, the height from the center of gravity of the soft rope to the top of the inclined plane is h2=L2, so the gravitational potential energy of the soft rope * * decreases mg(L2? L4) = L4mg。 So, C is correct.

D. Taking the system composed of soft rope and block as the research object, the gravitational potential energy of soft rope and block is reduced and converted into kinetic energy of block and soft rope and internal energy generated by friction between soft rope and inclined plane. According to the law of energy transformation and conservation, the decrease of gravitational potential energy of soft rope is less than the sum of the increase of kinetic energy and the work done to overcome friction. Therefore, d is correct.

So choose BCD.