题目内容
如图所示,质量M=4kg长为L=1Om的木板停放在光滑水平面上,另一不计长度质量m=1kg的木块以某一速度从右端滑上木板,木板与木块间的动摩擦因数
=0.8.若要使木板获得的速度不大于2m/S,木块的初速度V0应满足的条件为(g取10m/s2)( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250025011883613.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002501173446.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250025011883613.png)
A.![]() | B.![]() |
C.![]() | D.![]() |
BC
试题分析: 木块在木板上滑动时,木块的加速度a1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002501282559.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002501297672.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002501313465.png)
即v0t+
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002501329569.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002501344577.png)
若木块先减速后和木板一起匀速运动,根据动量守恒定律得:
mv0=(m+M)v,解得:v0=5v,因为v≤2m/s,所以v0≤10m/s。故选BC
![](http://thumb2018.1010pic.com/images/loading.gif)
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