题目内容
如图所示,质量为m的球置于斜面上,被一个竖直挡板挡住.现用一个力F拉斜面,使斜面在水平面上做加速度为a的匀加速直线运动,忽略一切摩擦,以下说法中错误的是
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241543490461929.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241543490461929.jpg)
A.若加速度a越小,竖直挡板对球的弹力越小 |
B.若加速度a逐渐变大,斜面对球的弹力不变 |
C.若加速度a逐渐变大,斜面对球的弹力减小 |
D.斜面和挡板对球的弹力的合力大于ma |
C
试题分析:分析小球受到的重mg、斜面的支持力
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154349061381.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154349186365.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154349061381.png)
小球受到的重mg、斜面的支持力
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154349061381.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154349186365.png)
则竖直方向有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154349061381.png)
,B正确;C错误;
∵mg和α不变,∴无论加速度如何变化,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154349061381.png)
水平方向有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154349186365.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154349061381.png)
A正确;斜面和挡板对球的弹力的合力即为竖直方向的
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154349061381.png)
故选答案C.
点评:本题结合力的正交分解考察牛顿第二定律,正确的分析受力与正确的分解力是关键.
![](http://thumb2018.1010pic.com/images/loading.gif)
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