题目内容
如图所示,质量 m =1kg的物块(可视为质点)在水平恒力F= 10N作用下,从水平面上 A点由静止开始运动,运动2s后再加一反向的水平恒力
=16N,问再经多少时间物块运动到B点,且A、B两点间的距离为
,已知物块与水平面间的动摩擦因数μ=0.20。(g=10m/s2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241151553972758.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155366206.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155381358.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241151553972758.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155412337.gif)
物块先作匀加速度直线运动,再作匀减速直线运动,最后做 反向的匀加速直线运动
在2s内,设加速度为
,末速度为
,位移
,
由由牛顿第二定律
得:![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155475568.gif)
有
;…………………… (1分)
…………………… (1分)
,代入数值得位移
=
…………………… (1分)
加反向力后,设物体作匀减速运动的加速度为
,由牛顿第二定律得![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155584628.gif)
代入数值得
=
,…………………… (2分)
设减速到零的时间为
,位移为
,由
得
,所以![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155678264.gif)
根据
,得
…………………… (1分)
设物体的反向加速的加速度为
,由牛顿第二定律得
,代入数值得
…………………… (2分)
设反向加速到A、B两点间的距离为19m的时间为
,这段时间内的位移为
,则
=
=
…………………… (1分)
由
,得
…………………… (1分)
所以再经时间
时A、B两点间的距离为
…………………… (2分)
在2s内,设加速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155428206.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155444200.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155459197.gif)
由由牛顿第二定律
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155459410.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155475568.gif)
有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155490899.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155506672.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155522471.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155459197.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155553363.gif)
加反向力后,设物体作匀减速运动的加速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155568210.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155584628.gif)
代入数值得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155568210.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155600397.gif)
设减速到零的时间为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155615199.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155631200.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155646422.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155678352.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155678264.gif)
根据
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155849456.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155849702.gif)
设物体的反向加速的加速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155865209.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155880616.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155896450.gif)
设反向加速到A、B两点间的距离为19m的时间为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155912201.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155927199.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155927199.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155990303.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115156005282.gif)
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115156021447.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115156036840.gif)
所以再经时间
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155412337.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115155381358.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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