A, the ball is moving in a straight line with uniform speed during sliding down and sliding up, so according to the average speed formula of uniform linear motion during sliding up and sliding down, the average speed during acceleration and deceleration is equal. V = v=v+v02, because the sliding distance is less than the upward sliding distance, the downward sliding time of the small ring is less than the upward sliding time of the small ring, and because the speed changes are the same when sliding downward and upward, the acceleration is a=△v△t, and the speed of the small ring when sliding downward is known.

B, as shown in the figure: the small ring is affected by gravity, supporting force and sliding friction on the rod. From the meaning of the question, f = μ n = μ gcos θ, because θ > α, the friction force when sliding downward is smaller than that when sliding upward, so B is correct;

C, when the small ring slides down, it makes uniform acceleration motion, the initial velocity is 0, the relationship between displacement and velocity satisfies v2=2ax: v=2ax, and the velocity is proportional to the square root of displacement, so c is wrong;

D. The work done by forces other than gravity and elasticity is equal to the change of mechanical energy of the small ring, so the mechanical energy when the ball slides down is equal to E=E0-fx. Because the friction force when it slides down is less than that when it slides up, the slope of the image when it slides down is less than that when it slides up, so D is wrong.

So: ab.