题目内容
如图所示,足够长的光滑斜面倾角θ=30°,一个带正电、电量为q的物体停在斜面底端B。现在加上一个沿斜面向上的场强为E的匀强电场,在物体运动到A点时撤销电场,那么:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250020063483737.jpg)
(1)若已知BA距离x、物体质量m,则物体回到B点时速度大小多少?
(2)若已知物体在斜面上运动的总时间是加电场时间的2倍,则物体的质量m是多少?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250020063483737.jpg)
(1)若已知BA距离x、物体质量m,则物体回到B点时速度大小多少?
(2)若已知物体在斜面上运动的总时间是加电场时间的2倍,则物体的质量m是多少?
(1)
;(2)
。(3)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006363751.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006379781.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250020063943727.jpg)
试题分析:
(1)物体由B运动再返回的过程中,重力做功为零,电场力做功为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006410484.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006441815.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006441829.png)
由于回到原来位置,即位移为零,则:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006457750.png)
得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006488483.png)
(2)设加电场时的时间为t1,该过程末速度大小为v1,返回到原点速度大小为v2,根据题意,整个过程如图所示。那么两过程加速度大小分别为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006504533.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006519662.png)
可见
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006519662.png)
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006550489.png)
再对物体在斜面上受力分析:
有电场时受到沿斜面向上的电场力、重力、支持力,根据牛顿第二定律,
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006566967.png)
没有电场时受到重力、支持力,根据牛顿第二定律,
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006582833.png)
由于
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006550489.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006379781.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目