(1) The surface of the inclined wedge A is smooth (let the inclination angle of the inclined plane be θ, the mass of A is mA, and the mass of B is mB).

A. If F 1 and F2 are removed at the same time, the object will also move in a straight line with uniform acceleration along the inclined plane under the action of the downward component of its gravity. Therefore, A is correct; ?

B if F 1 is removed, the external force moving relative to the ground is FN2sinθ=mB gcosθsinθ, and the direction is to the right, as shown in figure 1. Because MB GCOSθsinθ fcos θ, and because f=μFN, Fn sin θ > μ fn cos θ, that is, μ < tan θ.

A. recover F 1 and F2 at the same time. As can be seen from the above analysis, mbgsin θ > micron bgcos θ. Therefore, the resultant force on object B is downward along the inclined wedge, and the acceleration direction must also be downward along the inclined wedge. Therefore, A is correct;

B. If F 1 is removed, and the object B is still moving downward, then N=mgcosθ, and f = μ n. It is assumed that the direction of the friction force fA on A is left.

Nsinθ=fcosθ+fA, then: FA = Nsin θ-μ nos θ = n (sin θ-μ cos θ) > 0, so the wedge A tends to move to the right relative to the ground, and the direction of friction is left, so B is wrong;

C, and because the existence of F2 has no effect on the magnitude of ground friction on the inclined wedge, the magnitude and direction of friction on the inclined wedge A remain unchanged after F2 is removed. So, c is wrong;

D, according to the above judgment, so d is correct;

Therefore, in the case of rough surface of inclined wedge A, the correct option of this question is still AD.

So choose ad.