题目内容
如图所示,“
”型木块放在光滑水平地面上,木块水平表面AB粗糙,光滑表面BC且与水平面夹角为θ=37°.木块右侧与竖直墙壁之间连接着一个力传感器,当力传感器受压时,其示数为正值;当力传感器被拉时,其示数为负值.一个可视为质点的滑块从C点由静止开始下滑,运动过程中,传感器记录到的力和时间的关系如图所示.已知sin37°=0.6,cos37°=0.8,g取10m/s2.求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241526585696462.jpg)
(1)斜面BC的长度;
(2)滑块的质量;
(3)运动过程中滑块克服摩擦力做的功.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152658507217.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241526585696462.jpg)
(1)斜面BC的长度;
(2)滑块的质量;
(3)运动过程中滑块克服摩擦力做的功.
(1)3m (2) 2.5kg (3)40J
试题分析:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082415265860113667.png)
(1)分析滑块受力,由牛顿第二定律得:
得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152658663848.png)
通过图像可知滑块在斜面上运动时间为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152658725415.png)
由运动学公式得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152658866767.png)
(2)滑块对斜面的压力为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152658959800.png)
木板对传感器的压力为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152659037722.png)
由图像可知:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152659100564.png)
解得:m=2.5kg(说明:如果从系统考虑,答案正确得满分)
(3)滑块滑到B点的速度为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152659537720.png)
由图像可知:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152659599524.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152659661475.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152659771846.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152659880909.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152700114715.png)
点评:这类问题要分阶段研究,弄清每个阶段的受力情况及运动情况,根据各自的物理规律求解。
![](http://thumb2018.1010pic.com/images/loading.gif)
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