题目内容
如图所示,一个质量为0.6kg的小球以某一初速度从P点水平抛出,恰好从光滑圆弧AB的A点的切线方向进入圆弧(不计空气阻力,进入圆弧时无机械能损失),并从最低点B通过一段光滑小圆弧滑上另一粗糙斜面CD。已知圆弧AB的半径R=0.9m,θ=600,B在O点正下方,斜面足够长,动摩擦因数u=0.5,斜面倾角为370,小球从p到达A点时的速度为4m/s。(g取10m/s2,cos37°=0.8,sin37°=0.6)问:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250024456824910.jpg)
(1)P点与A点的水平距离和竖直高度
(2)小球在斜面上滑行的总路程
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250024456824910.jpg)
(1)P点与A点的水平距离和竖直高度
(2)小球在斜面上滑行的总路程
(1) x=
m;y=0.6m;(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002445697768.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002445697474.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002445697768.png)
试题分析: (1)A点的水平分速度为:vAx= vAcos60°=2m/s,
竖直分速度为:vAy= vAsin60°=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002445713423.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002445729494.png)
P点与A点的水平距离x= vAxt=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002445697474.png)
P点与A点的竖直高度y=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002445760550.png)
(2)设小球斜面向上运动的距离为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002445775266.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250024457912244.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002445807574.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250024458222082.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002445838849.png)
故小球不能过A点,只能来回摆动,最后停在B点,由能量守恒定律
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250024458531779.png)
得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002445697768.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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