题目内容
如图所示,绝缘水平板面上,相距为L的A?B两个点分别固定着等量正点电荷.O为AB连线的中点,C?D是AB连线上的两点,AC=CO=OD=OB=1/4L.一质量为m?电量为+q的小滑块(可视为质点)以初动能E0从C点出发,沿直线AB向D运动,滑动第一次经过O点时的动能为nE0(n>1),到达D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/20140824/201408241424534553838.jpg)
(1)小滑块与水平板面之间的动摩擦因数μ;
(2)OD两点间的电势差UOD;
(3)小滑块运动的总路程s.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241424534553838.jpg)
(1)小滑块与水平板面之间的动摩擦因数μ;
(2)OD两点间的电势差UOD;
(3)小滑块运动的总路程s.
(1)2E0/Lmg (2)
(1-2n) (3)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824142453845692.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824142453470546.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824142453845692.png)
(1)根据动能定理.
C―→O:W电+Wf=nE0-E0①
O―→D:W电′+Wf=0-nE0②
Wf=-μmg
③
根据题意W电=-W′电④
联立①②③④解得μ=2E0/Lmg.⑤
(2)根据动能定理,O到D过程有
qUOD-μmg
=O-nE0⑥
解⑤⑥得U0D=
(1-2n).⑦
(3)由初始C点至最终停止于O点,根据动能定理有qUCO-μmgS=0-E0⑧
UCO=-UOD⑨
联立⑤⑦⑧⑨解得总路程S=
.⑩
C―→O:W电+Wf=nE0-E0①
O―→D:W电′+Wf=0-nE0②
Wf=-μmg
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824142453891372.png)
根据题意W电=-W′电④
联立①②③④解得μ=2E0/Lmg.⑤
(2)根据动能定理,O到D过程有
qUOD-μmg
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824142453891372.png)
解⑤⑥得U0D=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824142453470546.png)
(3)由初始C点至最终停止于O点,根据动能定理有qUCO-μmgS=0-E0⑧
UCO=-UOD⑨
联立⑤⑦⑧⑨解得总路程S=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824142453845692.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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