题目内容
一质量为m的物体,沿半径为R的向下凹的圆形轨道滑行,如图所示,经过最低点的速度为v,物体与轨道之间的动摩檫因数为μ,则它在最低点时受到的摩檫力为( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241338039681715.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241338039681715.png)
A.μmg | B.μmv2/R |
C.μm(g+v2/R) | D.μm(g-v2/R) |
C
在最低点时
,摩擦力f=μF=μm(g+v2/R) ,C 对;
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241338040141025.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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题目内容
A.μmg | B.μmv2/R |
C.μm(g+v2/R) | D.μm(g-v2/R) |