题目内容
(选考题)(13分)如图所示,由细管道组成的竖直轨道,其圆形部分半径分别是R和
,质量为m的小球通过这段轨道时,在A处刚好对管壁无压力,在B点处对内轨道的压力为
.求由A运动到B的过程
中摩擦力对小球做的功.![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241218554286101.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121855382231.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121855397379.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412185541385.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241218554286101.jpg)
-
mgR
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121855444243.gif)
小球由A运动到B的过程中做
变速圆周运动,摩擦力既是变力又是未知力,故考虑用动能定理间接求此力做的功.
在A点用牛顿第二定律:mg=m![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121855506372.gif)
同理在B点为:mg-FN=m![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121855522413.gif)
由已知条件:FN=
mg
在A→B过程中对小球根据动能定理得:
Wf +mgR=
mvB2-
mvA2
联立以上各式解得:Wf =-
mgR.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412185549185.gif)
在A点用牛顿第二定律:mg=m
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121855506372.gif)
同理在B点为:mg-FN=m
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121855522413.gif)
由已知条件:FN=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121855538210.gif)
在A→B过程中对小球根据动能定理得:
Wf +mgR=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121855538210.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121855538210.gif)
联立以上各式解得:Wf =-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121855444243.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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