According to the centripetal force formula FN-mg=mv2R, fn = mv2r+mg = 2mgrh+mg is obtained;
According to PQ curve, when mg=5, m=0.5? kg。
From the image, we can know that for 2mgR= 10, R= 1? M. obviously when H=0.2? M corresponds to point d in the figure,
So cos? θ= 1? 0.2 1=0.8,θ=37 .
(2) If the block slides down the inclined plane, it is concluded from the kinetic energy theorem: mgH-μmgcos? θ(H？ 0.2)sinθ= 12mv2
Mv2 = 2 mgh-83 μ g (h-0.2) was obtained.
From the centripetal force formula, fn = mv2r+mg = 2mg? 83μmgRH+ 1.63μmg+mg
From the centripetal force formula FN-mg=mv2R, FN=mv2R+mg=2mg? +f(83μmg，R)H+ 1.63μmg+mg
Combining with the homogeneous curve, we know that 1.63μ g+mg = 5.8, and the solution is μ = 0.3.
(3) If the speed of the stone sliding down the slope to the highest point is V,
From kinetic energy theorem: mg(H-2R)-μmgcos? θ(H？ 0.2)sinθ= 12mv2？ ( 1)
Let the square just reach the highest point: centripetal force formula: mg=mv2R? (2)
It can be obtained from the formula (1)(2): h =15.1m.
Answer: (1) Find the mass m of the small block; The radius r of the circular orbit and the central angle θ corresponding to the orbit DC are 37;
(2) The dynamic friction coefficient between the small block and the inclined surface AD is 0.3.
(3) If the small block can move to the highest point E of the circular orbit, it should be released from the ground at a height of15.1m..