At the moment of (1) line shearing, Newton's second law is applied to the guide rail:

Mgsin37 -f=Ma

F = μ n f=μN=μmgcos37。

Solution: a = gsin37-μ mmgcos37.

Substitute into the solution, a=2.8m/s2?

(2) For the sliding process of the guide rail, Newton's second law is used:

Mgsin37 -f-FA=Ma

Where f = μ (mg cos 37-fa) and FA=B2L2vR.

Simultaneous: a = mgsin37? μ(mgcso37？ B2L2vR)？ B2L2RvM

= gsin 37-μmmgcos 37-B2L2vR( 1？ μ)

When a=0, the maximum speed of the guide rail is:

vm=(Mgsin37？ μmgcos37 )RB2L2( 1？ μ)= 5.6 m/s

(3) When the sliding distance d of the guide rail reaches the maximum speed, there are:

Q =。 I△ t =△φ r = BLDR Solution: d=6m? Apply the law of conservation of energy to the system;

Mgd sin 37 = 12mv2+△ e loss.

Substituting the data, the result is: △E loss = 20.32 J.

Answer: (1) When the thin wire is cut short, the acceleration of abcd motion of the guide rail is 2.8m/S2；; ;

(2) The maximum speed of abcd motion of guide rail is 5.6m/s;

(3) In this process, the mechanical energy lost by the system is 20.32 J. 。