题目内容
如图所示,一固定的楔形木块,其斜面的倾角θ=30°,另一边与地面垂直,顶上有一定滑轮.一柔软的细线跨过定滑轮,两端分别与物块A和B连结,A的质量为4m,B的质量为m,开始时将B按在地面上不动,然后放开手,让A沿斜面下滑而B上升.物块A与斜面间无摩擦.设当A沿斜面下滑S 距离后,细线突然断了.求物块B上升离地的最大高度H. ![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242036476407755.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242036476407755.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824203648529544.png)
试题分析:对系统由机械能守恒定律
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242036494971283.png)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824203650261714.png)
细线断后,B做竖直上抛运动,又上升了
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824203651431313.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824203652617795.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824203648529544.png)
点评:本题难度较小,处理连接体问题通常利用机械能守恒定律,此题注意两物体上升或下降高度并不想等
![](http://thumb2018.1010pic.com/images/loading.gif)
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