题目内容
(10分)如图所示,水平地面的B点右侧有一圆形挡板。圆的半径R=4m,B为圆心,BC连线与竖直方向夹角为37o.滑块静止在水平地面上的A点,AB间距L=4.5m.现用水平拉力F=18N沿AB方向拉滑块,持续作用一段距离后撤去,滑块恰好落在圆形挡板的C点,已知滑块质量辨=2kg,与水平面间的动摩擦因数
=0.4,取g=10m/s2,sin37 o =0.6,cos37 o=0.8.求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250038351205921.jpg)
(1)拉力F作用的距离,
(2)滑块从A点运动到圆弧上C点所用的时间.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003835104301.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250038351205921.jpg)
(1)拉力F作用的距离,
(2)滑块从A点运动到圆弧上C点所用的时间.
(1)
(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003835151491.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003835135542.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003835151491.png)
试题分析:(1)滑块离开B点后做平抛运动,设其速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003835182330.png)
水平方向:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003835198764.png)
竖直方向:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003835213916.png)
在水平面拉力F作用的距离为x,根据动能定理则有
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250038352441052.png)
带入数据计算得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003835135542.png)
(2)滑块从A到C共分为两个阶段,第一个阶段匀加速直线运动,加速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003835463997.png)
根据位移公式有
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003835478870.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003835494416.png)
匀加速的末速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003835525717.png)
此后在摩擦力作用下匀减速即
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003835541798.png)
匀减速运动时间
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003835556789.png)
滑块从A点运动到圆弧上C点所用的时间
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003835588712.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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