题目内容
(9分)如图所示,在绝缘水平面上,相距为L的A、B两点处分别固定着两个等量正电荷。a、b是AB连线上两点,其中Aa=Bb=L/4,a、b两点电势相等,O为AB连线的中点。一质量为m带电量为+q的小滑块(可视为质点)以初动能E0从a点出发,沿AB直线向b运动,其中小滑块第一次经过O点时的动能为初动能的n倍(n>1),到达b点时动能恰好为零,小滑块最终停在O点,求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241502023712168.jpg)
(1)小滑块与水平面间的动摩擦因数μ;
(2)Ob两点间的电势差Uob;
(3)小滑块运动的总路程s。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241502023712168.jpg)
(1)小滑块与水平面间的动摩擦因数μ;
(2)Ob两点间的电势差Uob;
(3)小滑块运动的总路程s。
(1)
(2)
(3)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150202574604.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150202386693.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150202496851.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150202574604.png)
试题分析:(1)由Aa=Bb=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150202652368.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150202714507.png)
设小滑块与水平面间的摩擦力大小为f,对于滑块从a→b过程,由动能定理得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150202776980.png)
而f=μmg ③
由①——③式得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150203026796.png)
(2)对于滑块从O→b过程,由动能定理得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241502030881037.png)
由③——⑤式得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241502032131046.png)
(3)对于小滑块从a开始运动到最终在O点停下的整个过程,由动能定理得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150203260822.png)
而
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241502033221191.png)
由③——⑧式得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150203369709.png)
点评:难题。力学中的很多规律在电学中同样适用,在选择规律时,首先考虑功和能的原理,其次考虑牛顿定律,因为前者往往更简捷。此题注意摩擦力在整个过程中始终做负功,其所做功与运动的总路程S有直接关系。
![](http://thumb2018.1010pic.com/images/loading.gif)
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