题目内容
一可看作质点滑块从一平台右端以某一速度水平抛出,恰好到右下方倾角为
的斜面顶端时速度沿斜面方向并沿斜面运动到斜面底端。已知平台到斜面顶端的竖直高度
,斜面与滑块之间的摩擦因数为
,斜面顶端底端的竖直高度
,
求:
(1)滑块水平抛出的初速度大小![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174043005325.png)
(2)滑块从抛出到斜面底端的时间![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174043255267.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241740434891950.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174041820496.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174042022606.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174042288553.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174042490546.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174042740660.png)
(1)滑块水平抛出的初速度大小
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174043005325.png)
(2)滑块从抛出到斜面底端的时间
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174043255267.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241740434891950.jpg)
(1)
(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174043770428.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174043504702.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174043770428.png)
试题分析:(1)设从抛出到达斜面顶端的时间为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174044066292.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174044316351.png)
则由运动学公式得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174044550669.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174044768517.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174044986514.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174045252618.png)
因为到达斜面顶端的速度方向沿斜面向下,所以有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174045517706.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174043504702.png)
(2)设到达斜面顶端速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174046016265.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174046328283.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174046718329.png)
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174046968476.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174047264934.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174047560972.png)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174047779498.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174048028544.png)
所以总时间
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824174048403669.png)
点评:本题是平抛运动和匀加速运动的综合,关键抓住两个过程的联系:平抛的末速度等于匀加速运动的初速度.
![](http://thumb2018.1010pic.com/images/loading.gif)
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