题目内容
如图所示,在光滑水平面上放有质量为M = 3kg的长木板,在长木板的左端放有m = 1kg的小物体,小物体大小可忽略不计。小物块以某一初速度
匀减速运动。已知小物块与长木板表面动摩擦因数
,当小物块运动了t = 2.5s时,长木板被地面装置锁定,假设长木板足够长(g=10m/s2)
求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005149130624.png)
(1)小物块刚滑上长木板时,长木板的加速度大小?
(2)长木板被锁定时,小物块距长木板左端的距离?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005149099554.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005149114473.png)
求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005149130624.png)
(1)小物块刚滑上长木板时,长木板的加速度大小?
(2)长木板被锁定时,小物块距长木板左端的距离?
(1)
(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005149177359.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005149145449.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005149177359.png)
试题分析:(1)小物体冲上长木板后受重力、压力和摩擦力的作用做匀减速运动,长木板摩擦力的作用下做匀加速运动, 对小物体有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005149364759.png)
根据牛顿第二定律得:对长木板:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005149567592.png)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250051495821019.png)
(2)对小物体由牛顿第二定律得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005149598554.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005149613757.png)
设二者达到共同速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005149645265.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005149660319.png)
对小物体
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005149676576.png)
对长木板
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005149691503.png)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005149723646.png)
这说明小物体冲上长木板1s后二者再一起匀速运动1.5被锁定,所以长木板被锁定时,小物块距长木板左端的距离为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250051497381150.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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