A. From the figure, the acceleration of A at time t2 is zero and the speed is maximum. According to Newton's second law and Hooke's law, we get: mgsinθ=kx,
Then we get: x =mgsinθk, so A is wrong;
B. From the figure, A and B begin to separate at time t1. For A, according to Newton's second law: kx-mgsinθ=ma, then x=mgsinθ+mak, Therefore B is correct.
C. From the beginning to time t1, for AB as a whole, according to Newton's second law: F+kx-mgsinθ=2ma, F=mgsinθ+2ma-kx, x decreases, and F increases ; From time t1 to time t2, for B, according to Newton's second law: F-mgsinθ=ma, we get F=mgsinθ+ma. It can be seen that F does not change, so C is wrong.
D. From the above: A and B begin to separate at time t1...①
At the beginning: 2mgsinθ=kx0...②
From the beginning to time t1 , the potential energy released by the spring Ep=12kx20-12kx2…③
In the process from the beginning to t1, according to the kinetic energy theorem: WF+Ep-2mgsinθ (x0-x)=12mv21…④
2a(x0-x)=v12…⑤
The solution from ①②③④⑤ is: WF-Ep=-(mgsinθ?ma)2k, so the work done by the pulling force F is less than the potential energy released by the spring , so D is correct.
Hence the choice: BD