摘要:v0 = -------② vy = gt --------③

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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:

mghmv                                                       ①

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

Hgt2                                                                                                                                                     

v1t                                                                                                               

联立②③两式得:v1=4 m/s

设释放小球的高度为h1,则由mgh1mv

h1=0.8 m.

(3)由机械能守恒定律可得:mghmv2

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:

ygt2                                                                                                                                                      

xvt                                                                                                                      

tan 37°=                                                                                                         

vygt                                                                                                                     

vv2v                                                       ⑧

Ekmv                                                      ⑨

由④⑤⑥⑦⑧⑨式得:Ek=32.5h                                                                      

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)Ek=32.5h 图象见解析

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