题目内容
(14分)如图所示,质量mA=1kg的小物块以向右VA=4.0m/s的初速度滑上质量mB=1.0kg以向左初速度VB=5.0m/s的长木板,已知A、B之间的动摩擦因数μ1="0.20" ,B与地面之间的动摩擦因数μ2=0.40,整个过程中小物块并未从长木板上滑下,g取10 m/s2。则:
(1)求小物块A刚滑上长木板B时,小物块与长木板的加速度。
(2)求从小
物块A刚滑上长木板B到二者刚好相对静止时小物块的位移的大小。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241216016444414.jpg)
(1)求小物块A刚滑上长木板B时,小物块与长木板的加速度。
(2)求从小
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412160162972.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241216016444414.jpg)
(1)小木块加速度为2.0m/s2,方向向左;
长木板加速度为10m/s2,方向向右
(2)4m
长木板加速度为10m/s2,方向向右
(2)4m
(1)以A作为研究对象,由牛顿第二定律得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241216017381059.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241216017531144.jpg)
以B作为研究对象,由牛顿第二定律得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241216018781534.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241216018941046.jpg)
(2)由题意可知A、B都做减速运动,设B的速度先变为0时且所用时间为t1,
则对B:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241216019091037.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121601925654.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241216020501058.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121602065730.jpg)
此后A对B的摩擦力小于地面对B的摩擦力,所以B停止运动,A以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241216020811600.jpg)
由运动学公式:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121602284950.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121602299690.jpg)
![](http://thumb2018.1010pic.com/images/loading.gif)
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