题目内容
(12分)如图所示,水平桌面上有一轻弹簧,左端固定在A点,自然状态时其右端位于B点。水平桌面右侧有一竖直放置的光滑轨道MNP,其形状为半径R=0.8m的圆环剪去了左上角135°的圆弧,MN为其竖直直径,P点到桌面的竖直距离也是R、水平距离是2R。用质量m=0.4kg的物块将弹簧缓慢压缩到C点,释放后物块过B点后其位移与时间的关系为
,物块飞离桌边缘D点后由P点沿切线落入圆轨道。g=10m/s2,求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250019551473098.jpg)
(1)BD间的水平距离;
(2)滑块运动到N处时对轨道的压力;
(3)判断m能否沿圆轨道到达M点(说明理由)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001955100527.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250019551473098.jpg)
(1)BD间的水平距离;
(2)滑块运动到N处时对轨道的压力;
(3)判断m能否沿圆轨道到达M点(说明理由)
试题分析:
(1)设物块由D点以初速度v0做平抛,落到P点时其竖直分速度为vy,
竖直方向上
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001955162641.png)
水平方向上
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001955178578.png)
得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001955194632.png)
由表达式
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001955100527.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001955225636.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001955240622.png)
则BD间位移x满足:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250019552561065.png)
(2)mg(R+Rcos45°)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001955272856.png)
FN-mg="m" VN2/R 得FN=(20+4
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001955287344.png)
(3)若物块能沿轨道到达M点,其速度为vm
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250019553031334.png)
轨道对物块的压力为FN,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250019553181010.png)
解得FN=(1-2)mg<0
即物块不能到达M点 2分
![](http://thumb2018.1010pic.com/images/loading.gif)
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