题目内容
如图所示,OD是一水平面,AB为一斜面,物体经过B处时无能量损失,一质点由A点静止释放,沿斜面AB滑下,最后停在D点,若斜面改为AC(仅倾角变化),仍从A点由静止释放,则最终停在水平面OD上的(设各处动摩擦因数相同)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242125004941975.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242125004941975.jpg)
A.D点右侧 | B.D点左侧 |
C.D点 | D.无法确定 |
C
试题分析:过程中摩擦力做负功,重力做正功,故根据动能定理可得:沿AB运动时:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242125012891452.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242125022411923.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824212503146826.png)
所以物体静止时的位置与斜面的倾斜程度无关,沿AC运动时,仍静止于D点,
故选C
点评:本题的关键是推出重力做功与摩擦力做功关系为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824212503146826.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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