(2014?Linyi simulation) in the angle of inclination of θ on the smooth inclined plane there are two with a light spring connected to the object A, B, their mass are m, the spring strength system

A, at the beginning, the spring is in compression, the pressure is equal to the sliding force of gravity of the object A, according to Hooke's law, there are:

mgsinθ=kx1

Solution:

x1=

mgsinθ

k

When object B is just about to move away from C, the tension of the spring is equal to the downward sliding force of the gravitational force of object B. According to Hooke's law, the tension is equal to the downward sliding force of object B. The component force, according to Hooke's law, has;

mgsinθ=kx2

Solve:

x2=

mgsinθ

k

Therefore, the distance traveled by block A is: △x = x1 + x2 =

2mgsinθ

k

, therefore A is correct;

B, at this time object A by tension, gravity, support and spring tension, according to Newton's second law, there are:

F-mgsinθ-T=ma

Spring tension is equal to the downward sliding force of gravity of the object B, is:

T=mgsinθ

Therefore: a=

F

m

?2gsinθ, so B is wrong;

C, D, the work done by the tension force F is equal to the increase in mechanical energy of object A, object B and the spring system as:

W=mg?△xsinθ+

1

2

mv2+EP bomb, so C is wrong, D is wrong;

Therefore, the choice is:

The choice is:

A. p>