Cos photographer Qingdao

(1) for m, it is obtained from the equilibrium condition: mgsinθ=μ2mgcosθ,

The solution is: μ 2 = tan θ;

(2) for m, let its maximum acceleration be a,

From the equilibrium condition: FNcosθ=mg+μ2FNsinθ,

Newton's second law: FNsinθ+μ2FNcosθ=ma,

The solution is: a=2gsinθcosθ? tanθsinθ，

For the system composed of m and m, it is obtained by Newton's second law:

F-μ 1(M+m)g=(M+m)a，

The solution is: f = μ1(m+m) g+2 (m+m) gsinθ cos θ? tanθsinθ；

(3) For the system composed of m and m, we can get the following from the dynamics theorem:

FD-μ 1(M+M)GD = 12(M+M)v2-0，

The solution is: v=2gdsinθcosθ? tanθsinθ，

M for flat throwing, vertical direction: h= 12gt2, horizontal direction: xP=vt-htanθ,

Solution: xP=22hdsinθcosθ? tanθsinθ-htanθ；

Answer: (1) The coefficient of dynamic friction between the small block and the inclined plane μ 2 = tan θ;

(2) Make the falling point P of the block on the ground farthest from the obstacle Q, and the horizontal thrust is μ1(m+m) g+2 (m+m) gsinθ cos θ? tanθsinθ；

(3) The farthest distance from the landing point P of small objects on the ground to the obstacle Q is 22hdsinθcosθ? tanθsinθ-htanθ。