题目内容
如图所示,光滑绝缘半球槽的半径为R,处在水平向右的匀强电场中,一质量为m的带电小球从槽的右端A处无初速沿轨道滑下,滑到最低位置B时,对轨道的压力为2mg.求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241141042521412.jpg)
(1)小球所受电场力的大小和方向;
(2)带电小球在滑动过程中的最大速度.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241141042521412.jpg)
(1)小球所受电场力的大小和方向;
(2)带电小球在滑动过程中的最大速度.
(1)
mg,水平向右 (2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114104610438.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114104470225.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114104610438.gif)
(1)设小球运动到最低位置B时速度为v,此时有
FN-mg=
①
小球从A处沿槽滑到最低位置B的过程中,根据动能定理有
mgR+W电=
mv2 ②
由①②式两式解得W电=-
mgR,电场力做负功,则电场力方向必水平向右.
设电场力大小为F,由W电=-FR得F=
mg.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241141047351861.jpg)
(2)小球在滑动过程中速度最大的条件是小球沿轨道运动至某位置时切向合力为零.设此时小球与圆心连线与竖直方向的夹角为θ,如图mgsinθ=Fcosθ ③
由③得:tanθ=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114104470225.gif)
小球由A处到最大速度位置的过程中
mgRcosθ-
mgR(1-sinθ)=
(mvm2-0)
得:vm=
.
FN-mg=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114104673414.gif)
小球从A处沿槽滑到最低位置B的过程中,根据动能定理有
mgR+W电=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114104470225.gif)
由①②式两式解得W电=-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114104470225.gif)
设电场力大小为F,由W电=-FR得F=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114104470225.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241141047351861.jpg)
(2)小球在滑动过程中速度最大的条件是小球沿轨道运动至某位置时切向合力为零.设此时小球与圆心连线与竖直方向的夹角为θ,如图mgsinθ=Fcosθ ③
由③得:tanθ=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114104470225.gif)
小球由A处到最大速度位置的过程中
mgRcosθ-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114104470225.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114104470225.gif)
得:vm=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114104610438.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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