题目内容
如图所示.一个质量为m=10kg的物体, 由1/4光滑圆弧轨道上端从静止开始下滑, 然后滑上粗糙水平面向右滑动2.0m的距离而停止.已知轨道半径R=0.8m, g=10m/s2, 求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242108285032311.jpg)
①物体滑至轨道底端时的速度?
②物体与水平面间的动摩擦因数μ?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242108285032311.jpg)
①物体滑至轨道底端时的速度?
②物体与水平面间的动摩擦因数μ?
①
②µ=0.4
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824210829112836.png)
试题分析:(1)由机械能守恒得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824210830297782.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824210829112836.png)
(2)由动能定理有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824210831967865.png)
点评:本题难度较小,熟练掌握机械能守恒的条件,在水平面只有滑动摩擦力做功,由动能定理求解动摩擦因数
![](http://thumb2018.1010pic.com/images/loading.gif)
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