题目内容
如图所示,固定在竖直平面内倾角为
的直轨道AB,与倾角可调的足够长的直轨道BC顺滑连接。现将一质量
的小物块,从高为
处静止释放,沿轨道AB滑下,并滑上倾角也为
的轨道BC,所能达到的最大高度是
。若物块与两轨道间的动摩擦因数相同,不计空气阻力及连接处的能量损失。已知
,
,取g=10m/s2,求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241955375634554.jpg)
(1) 物块从释放到第一次速度为零的过程中,重力所做的功;
(2) 物块与轨道间的动摩擦因数
。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824195531697474.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824195532649590.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824195533367653.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824195533944377.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824195534583656.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824195535769641.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824195536627644.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241955375634554.jpg)
(1) 物块从释放到第一次速度为零的过程中,重力所做的功;
(2) 物块与轨道间的动摩擦因数
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824195537953301.png)
(1)0.3J(2)μ=0.25
试题分析:(1)物体下降了
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824195538468593.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824195539107747.png)
(2)根据动能定理可得由
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241955397941270.png)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241955403551170.png)
得μ=0.25(3分)
点评:在使用动能定理分析问题时,一定要注意过程中的始末状态
![](http://thumb2018.1010pic.com/images/loading.gif)
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