解析　當小木塊剛放到傳送帶上時μmg＝ma得a＝μg

設(shè)小木塊加速到v時運動的距離為L₀，由v²＝2aL₀得L₀＝

(1)當L≤L₀時，小木塊一直加速，v≥at得t≤或由L＝at²得t＝ 或由L≤vt得t≥，故C、D正確．

(2)當L>L₀時，小木塊先加速后勻速，加速階段有v＝at₁得t₁＝

勻速階段有L－L₀＝vt₂得t₂＝－

由t＝t₁＋t₂得t＝＋，故A正確．

答案　ACD

解析　(1)小球從曲面上滑下，只有重力做功，由機械能守恒定律知：

mgh＝mv                                                       ①

得v₀＝＝ m/s＝2 m/s.

(2)小球離開平臺后做平拋運動，小球正好落在木板的末端，則

H＝gt²                                                                                                                                                      ②

＝v₁t                                                                                                                ③

聯(lián)立②③兩式得：v₁＝4 m/s

設(shè)釋放小球的高度為h₁，則由mgh₁＝mv

得h₁＝＝0.8 m.

(3)由機械能守恒定律可得：mgh＝mv²

小球由離開平臺后做平拋運動，可看做水平方向的勻速直線運動和豎直方向的自由落體運動，則：

y＝gt²                                                                                                                                                       ④

x＝vt                                                                                                                       ⑤

tan 37°＝                                                                                                          ⑥

v_y＝gt                                                                                                                      ⑦

v＝v²＋v                                                       ⑧

E_k＝mv                                                      ⑨

由④⑤⑥⑦⑧⑨式得：E_k＝32.5h                                                                       ⑩

考慮到當h>0.8 m時小球不會落到斜面上，其圖象如圖所示

答案　(1)2 m/s　(2)0.8 m　(3)E_k＝32.5h　圖象見解析