题目内容
如图所示,AB为固定在竖直平面内的
光滑圆弧轨道,轨道的B点与水平地面相切,其半径为R.质量为m的小球由A点静止释放,求:
(1)小球滑到最低点B时,小球速度v的大小及小球对轨道的压力F压的大小;
(2)小球通过光滑的水平面BC滑上固定曲面,恰达最高点D,D到地面的高度为h(已知h<R),则小球在曲面上克服摩擦力所做的功Wf.![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241345276322520.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134527430303.png)
(1)小球滑到最低点B时,小球速度v的大小及小球对轨道的压力F压的大小;
(2)小球通过光滑的水平面BC滑上固定曲面,恰达最高点D,D到地面的高度为h(已知h<R),则小球在曲面上克服摩擦力所做的功Wf.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241345276322520.png)
(1)
(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134527726911.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134527664654.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134527726911.png)
解:⑴由动能定理得
,………………………………2分
解得
………………………………1分
由牛顿第二定律得
,………………………………2分
解得
………………………………1分
(2)由动能定理得
………………………………4分
解得
……………………2分
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134527898784.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134527929580.png)
由牛顿第二定律得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134527991903.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134527664654.png)
(2)由动能定理得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241345280691096.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134527726911.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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