题目内容
(15分)如图所示,质量为m的木板静止地放在光滑水平面上,质量为2m、可视为质点的木块以水平速度v0从左端滑上木板,木块与木板间的动摩擦因数为μ,木板足够长。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250008299942017.jpg)
⑴求木块和木板的加速度大小;
⑵求木块和木板速度相等所经历的时间及此时木块相对于木板的位移;
⑶若木板不是足够长,要使木块不从木板上滑落,求木板的最小长度。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250008299942017.jpg)
⑴求木块和木板的加速度大小;
⑵求木块和木板速度相等所经历的时间及此时木块相对于木板的位移;
⑶若木板不是足够长,要使木块不从木板上滑落,求木板的最小长度。
⑴a1=μg,a2=2μg;⑵t=
,Δs=
;⑶Lmin=![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000830025627.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000830009587.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000830025627.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000830025627.png)
试题分析:⑴对木块,根据牛顿第二定律有:2μmg=2ma1,解得木块的加速度大小为:a1=μg
对木板,根据牛顿第二定律有:2μmg=ma2,解得木块的加速度大小为:a2=2μg
⑵设经时间t两者速度相等,且为v,此过程中木块的位移为s1,木板的位移为s2,根据匀变速直线运动速度公式有:v=v0-a1t=a2t,解得:t=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000830009587.png)
根据匀变速直线运动位移公式有:s1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000830243580.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000830259470.png)
木块相对于木板的位移为:Δs=s1-s2,综合以上各式解得:Δs=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000830025627.png)
⑶由第⑵问的分析可知,当木板的长度L=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000830025627.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000830025627.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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