题目内容
(16分)水上滑梯可简化成如图所示的模型,斜槽AB和水平槽BC平滑连接,斜槽AB的竖直高度H=5.0m,倾角θ=37°。BC面与水面的距离h=0.80m,人与AB、BC间的摩擦均忽略不计。取重力加速度g=10m/s2,cos37°=0.8,sin37°=0.6。一同学从滑梯顶端A点无初速地自由滑下,求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250041094824258.png)
(1)该同学沿斜槽AB下滑时加速度的大小a;
(2)该同学滑到B点时速度的大小vB;
(3)从C点滑出至落到水面的过程中,该同学在水平方向位移的大小x。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250041094824258.png)
(1)该同学沿斜槽AB下滑时加速度的大小a;
(2)该同学滑到B点时速度的大小vB;
(3)从C点滑出至落到水面的过程中,该同学在水平方向位移的大小x。
(1)
(2)
(3)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004109529483.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004109497633.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004109513655.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004109529483.png)
试题分析:(1)该同学沿斜槽AB下滑过程中,受重力和斜槽的支持力,由牛顿第二定律
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004109544691.png)
可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004109497633.png)
(2)由A到B的过程中只有重力做功,设到达B点时速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004109591351.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004109607817.png)
可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004109513655.png)
(3)由B到C该同学做匀速直线运动,到达C点时的速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004109653736.png)
水平方向
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004109669468.png)
竖直方向
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004109700650.png)
可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004109529483.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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