解析　当小木块刚放到传送带上时μmg＝ma得a＝μg

设小木块加速到v时运动的距离为L₀，由v²＝2aL₀得L₀＝

(1)当L≤L₀时，小木块一直加速，v≥at得t≤或由L＝at²得t＝ 或由L≤vt得t≥，故C、D正确．

(2)当L>L₀时，小木块先加速后匀速，加速阶段有v＝at₁得t₁＝

匀速阶段有L－L₀＝vt₂得t₂＝－

由t＝t₁＋t₂得t＝＋，故A正确．

答案　ACD

解析　(1)小球从曲面上滑下，只有重力做功，由机械能守恒定律知：

mgh＝mv                                                       ①

得v₀＝＝ m/s＝2 m/s.

(2)小球离开平台后做平抛运动，小球正好落在木板的末端，则

H＝gt²                                                                                                                                                      ②

＝v₁t                                                                                                                ③

联立②③两式得：v₁＝4 m/s

设释放小球的高度为h₁，则由mgh₁＝mv

得h₁＝＝0.8 m.

(3)由机械能守恒定律可得：mgh＝mv²

小球由离开平台后做平抛运动，可看做水平方向的匀速直线运动和竖直方向的自由落体运动，则：

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

考虑到当h>0.8 m时小球不会落到斜面上，其图象如图所示

答案　(1)2 m/s　(2)0.8 m　(3)E_k＝32.5h　图象见解析