题目内容
(16分) 如图所示,一条轨道固定在竖直平面内,ab段水平且粗糙,动摩擦因数为μ.bcde段光滑,cde段是以O为圆心、半径为R的一小段圆弧.可视为质点的物块A和B紧靠在一起,静止于b处,A的质量是B的3倍.两物体在足够大的内力作用下突然分离,分别沿轨道向左、右运动.B到d点时速度沿水平方向,此时轨道对B的支持力大小等于B所受重力的3/4,重力加速度g,![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250003135472761.png)
求:(1)物块B在d点的速度大小;
(2)分离后B的速度;
(3)分离后A在ab段滑行的距离.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250003135472761.png)
求:(1)物块B在d点的速度大小;
(2)分离后B的速度;
(3)分离后A在ab段滑行的距离.
(1)
;(2)
;(3)
。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000313610707.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000313625776.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000313641641.png)
试题分析:(1) 由于在d点时轨道对B的支持力大小等于B所受重力的3/4,
在d点对B受力分析,并运用牛顿第二定律可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000313672987.png)
而
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000313703784.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000313719708.png)
(2) B物体由d→b的过程中,遵循机械能守恒定律,
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250003137341065.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000313750778.png)
(3) 分离的过程,A、B动量守恒,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000313781665.png)
又因为分离后A滑行过程中,摩擦力做负功,故由动能定理得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250003137971019.png)
解之得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000313797568.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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