题目内容
(12分)如图所示,在水平方向的匀强电场中有一表面光滑、与水平面成45°角的绝缘直杆AC,其下端(C端)距地面高度h=0.8 m.有一质量为500 g的带电小环套在直杆上,正以某一速度沿杆匀速下滑,小环离开杆后正好通过C端的正下方P点.(g取10 m/s2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250008113524832.jpg)
求:(1)小环离开直杆后运动的加速度大小和方向;
(2)小环在直杆上匀速运动时速度的大小;
(3)小环运动到P点的动能.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250008113524832.jpg)
求:(1)小环离开直杆后运动的加速度大小和方向;
(2)小环在直杆上匀速运动时速度的大小;
(3)小环运动到P点的动能.
(1)14.1m/s2;垂直于杆斜向右下方(2)2m/s (3)5J
试题分析:(1)小环在直杆上的受力情况如图所示.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250008113832700.jpg)
由平衡条件得:mgsin45°=Eqcos45° ①
得: mg=Eq,
离开直杆后,只受mg、Eq作用,则:
F合=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000811398344.png)
a=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000811398344.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000811398344.png)
方向与杆垂直斜向右下方.
(2)设小环在直杆上运动的速度为v0,离杆后t秒到达P点,则:
竖直方向:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250008114611063.png)
水平方向:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250008114761150.png)
联立解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000811492921.png)
(3)由动能定理得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250008115231011.png)
代入数据解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000811539556.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目