题目内容
如右图所示,长为L的长木板水平放置,在木板的A端放置一个质量为m的小物块.现缓慢地抬高A端,使木板以左端为轴转动,当木板转到与水平面的夹角为α时小物块开始滑动,此时停止转动木板,小物块滑到底端的速度为v,则在整个过程中( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242139227701729.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242139227701729.jpg)
A.支持力对物块做功为0 |
B.支持力对小物块做功为![]() |
C.摩擦力对小物块做功为![]() |
D.滑动摩擦力对小物块做功为![]() |
B
试题分析:物块在缓慢提高过程中,由动能定理可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824213926280898.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824213927294809.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242139280581198.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824213928807883.png)
故选B
点评:当力是恒定时,除可由力与力的方向位移求出功外,还可以由动能定理来确定;当力是变化时,则只能由动能定理来求出力所做的功.
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目