题目内容
如图所示,足够长光滑绝缘斜面与水平面间的夹角为α=37°,处于水平方向的匀强电场和匀强磁场中,电场强度E="50" V/m,方向水平向左,磁场方向垂直纸面向外;一个电荷量q=+4.0×10-2C,质量m="0.40" kg的光滑小球,以初速度v0="20" m/s从斜面底端向上滑,然后又下滑,共经过t=3s脱离斜面。取g=10m/s2,sin37°=0.6,cos37°=0.8。求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250052525133235.jpg)
(1)小球上滑过程的加速度a1大小及时间t1;
(2)磁场的磁感应强度B。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250052525133235.jpg)
(1)小球上滑过程的加速度a1大小及时间t1;
(2)磁场的磁感应强度B。
(1)
,t1=2s(2)B=5T
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005252528587.png)
试题分析:(1)上滑过程,小球沿斜面向上做减速运动,受力分析如图,由牛顿第二定律和运动学规律得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250052525281988.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005252560925.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005252575522.png)
联解①②代入数据得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005252528587.png)
t1=2s ………………④
(2)小球下滑过程受力分析如图,在离开斜面前小球做匀加速直线运动,设运动时间为t2,脱离斜面时的速度为v2,由牛顿第二定律和运动学规律有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250052528092028.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005252825950.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250052528401043.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005252856405.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005252872471.png)
联解⑤⑥⑦⑧代入数据得:
B=5T ………………⑨
![](http://thumb2018.1010pic.com/images/loading.gif)
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