题目内容
如图所示,光滑的水平面上放着一块木板,木板处于静止状态,其质量M=2.0kg。质量m="1.0" kg的小物块(可视为质点)放在木板的最右端。现对木板施加一个水平向右的恒力F,使木板与小物块发生相对滑动。已知F=6N,小物块与木板之间的动摩擦因数μ=0.10,g取10m/s
。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250022021473203.jpg)
(1)求木板开始运动时的加速度大小;
(2)在F作用1s后将其撤去,为使小物块不脱离木板,木板至少多长?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002202084242.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250022021473203.jpg)
(1)求木板开始运动时的加速度大小;
(2)在F作用1s后将其撤去,为使小物块不脱离木板,木板至少多长?
(1) 2.5m/s2 (2) 1.5m
试题分析:(1)长木板水平方向的受力情况如答图1所示,木板向右做匀加速直线运动,设木板的加速度为a1
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250022021622120.jpg)
根据牛顿第二定律有:
F-μmg=Ma1 2分
a1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002202287780.png)
(2)t1=1s时撤去力F,之后木板向右做匀减速直线运动,设木板的加速度大小为a1
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250022023031102.png)
物块一直向右做匀加速直线运动,设物块的加速度为a2
a2=μg=1.0m/s2 1分
设再经过时间t2物块滑到木板的最左端,两者的速度相等,即
a1t1-a
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002202318245.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250022023493301.jpg)
木板的位移
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250022023651094.png)
物块的位移
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002202381875.png)
由答图2可知
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002202381504.png)
解得L=1.5m 1分
![](http://thumb2018.1010pic.com/images/loading.gif)
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