Due to the elasticity of the spring, A is N 1 upward along the slope, N2 upward due to the elasticity of the slope, and mg downward due to gravity.
Divide the elastic force into horizontal and vertical directions and draw your own picture. I didn't know how to type the angle, so I used a instead.
Then N2 Sina+n1COSA = mg-1.
N2cosa-N 1sina=ma - 2
At first, a = 0, because everything is still.
After a given force F, as long as there is F, it must be a>0, then N2 in Formula 2 will inevitably increase, and n 1 in Formula1will inevitably decrease.
Because the spring is compressed when it starts to rest, the expansion amount x, N 1 = KX, where x is smaller and the spring is elongated.
When f continues to increase, N 1 will decrease to 0. Then a will only have the elasticity and gravity of the slope.
At this time, because A can only have the elasticity and gravity of the inclined plane, then B and A are the same, so the pressure of the baffle will definitely be 0.
There is no doubt about it. So BD is right.