题目内容
如图所示,带正电的小球套在绝缘的长直杆上,球与杆之间的动摩擦因素为μ,匀强电场水平向右,将杆与竖直线分别成不同角度固定放置,让小球由静止释放后都沿杆运动距离s,当杆与竖直线的夹角为θ时,小球获得的末动能最大为E.现将杆竖直固定放置,为使小球由静止释放后获得的动能与前面的动能E相等,小球运动的距离
应是多少?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241201078091538.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120107794929.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241201078091538.gif)
1/(cosθ-μsinθ)
动能最大有:Wf=0 mgscosθ+Eqssinθ=Ek
竖直时:mg
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120107825926.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120107825926.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120107825926.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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