(1) The slider makes uniform deceleration motion on the inclined plane. v = υ 0+υ 2 = 9m/s，t = s.v = 0.25s。

(2) The velocity of the block moving at the top of the slope is υ t..

From bottom to top: υt2? υ02=2aX

Namely: υt? 2? 102=2a×4.5

From bottom to midpoint: υ 2? υ02=2aS？

Namely: 82- 102=2a×2.25.

The solution is a =-8m/S2 υ t = 28m/s.

The motion of the block is divided into two stages: one is the uniform deceleration motion along the inclined plane with an acceleration of 8m/s2, and the direction is downward along the inclined plane; Second, projectile motion, the acceleration is gravity acceleration g= 10m/s2, and the direction is vertical downward.

(3) Find the sliding friction coefficient first? The block moves uniformly along the inclined plane: mgsinθ+μ gcosθ = maμ = 0.25.

Analyze the force on the inclined plane (as shown in the figure) and establish the coordinate system. In the x direction:

f =(μmgcosθ)cosθ+(mgcosθ)sinθ= 12.8n

So the answer is: the time for the (1) box to move from the bottom to the midpoint is t = 0.25 s.

(2) The movement of the pulley is divided into two stages: one is the uniform deceleration movement along the inclined plane with an acceleration of 8m/s2, and the direction is downward along the inclined plane; Second, projectile motion, the acceleration is gravity acceleration g= 10m/s2, and the direction is vertical downward.

(3) When the block moves on the inclined plane, the horizontal ground friction on the inclined plane is 12.8N, and the direction is horizontal to the left.