题目内容
(10分)如图所示,质量为m=1kg的滑块,以υ0=5m/s的水平初速度滑上静止在光滑水平面的平板小车,若小车质量M=4kg,平板小车足够长,滑块在平板小车上滑移1s后相对小车静止。求:(g取10m/s2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250039571291240.png)
(1)滑块与平板小车之间的滑动摩擦系数μ; (2)此时小车在地面上滑行的位移?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250039571291240.png)
(1)滑块与平板小车之间的滑动摩擦系数μ; (2)此时小车在地面上滑行的位移?
(1)μ=0.4 (2)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003957145609.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003957145609.png)
试题分析:(1)m滑上平板小车到与平板小车相对静止,速度为v1,
据动量守恒定律:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003957160819.png)
对m据动量定理:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003957176759.png)
代入得μ=0.4 (2分)
(2)对M据动能定理有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250039571911043.png)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003957207608.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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