题目内容
如图所示,一个横截面积为S的圆筒型容器竖直放置,金属圆板A的上表面是水平的,下表面是倾斜的,下表面与水平面的夹角为θ,圆板的质量为M,不计圆板A与容器内壁之间的摩擦,若大气压强为P0,则被圆板封闭在容器中气体的压强p等于( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152244024992.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152244024992.png)
A.![]() | B.![]() |
C.![]() | D.![]() |
D
【错解分析】错解一:因为圆板下表面是倾斜的,重力产生的压强等于
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152244788586.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152244851338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152245053664.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152245209765.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152245350958.png)
错解二:重力产生的压强应该为重力的分力
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152245490665.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152245615811.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152245818581.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152245818581.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152245958738.png)
错解三:大气压p0可以向各个方向传递,所以气体压强里应包括p0,而重力的合力
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152246021628.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152245053664.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241522461301141.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152246192346.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152246457827.png)
【错解原因】重力产生的压强,压力都应该是垂直于接触面方向,所以重力产生压强应是重力的分力
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152245490665.png)
错解二虽然注意到重力的分力
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152245490665.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152244851338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152245053664.png)
错解三在分解重力时错了,重力的一个分力应是
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152245490665.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152246021628.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241522471281418.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241522472228734.png)
【正解】以金属圆板A为对象,分析其受力情况,从受力图可知,圆板A受竖直向下的力有重力Mg、大气压力p0S,竖直向上的力为其他压力
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152247300406.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152247534574.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152245053664.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241522476901177.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241522477521548.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152247846517.png)
正确答案应为D。
【点评】正如本题的“分析解答”中所做的那样,确定被活塞封闭的气体的压强的一般方法是:以活塞为研究对象;分析活塞的受力情况;概括活塞的运动情况(通常为静止状态),列出活塞的受力方程(通常为受力平衡方程);通过解这个方程便可确定出气体的压强。
![](http://thumb2018.1010pic.com/images/loading.gif)
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