题目内容
一根均匀的米尺,质量为0.2kg,放在水平桌面上,它与桌面间的动摩擦因数为0.16,
长度露在桌外,现有一水平拉力沿着米尺方向作用在米尺上作用1s,为使米尺能从桌边落下,外力F至少应为多少?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114804891221.gif)
F至少为0.4N
取米尺为研究对象,设从开始运动到停止运动用的时间为
位移为
,拉力停止作用对尺子的运动速度为
,则由动量定理可知。
对全过程有
①
拉力作用时间段
②
由运动学知识
③
联立①②可知![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114805172866.gif)
欲使尺子从水平桌面落下必使尺子的重心出桌子边缘。
∴
故有
代入数字可知
化简得![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114805234512.gif)
解得
故 F至少为0.4N
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114804906185.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114805031182.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114805047185.gif)
对全过程有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114805078537.gif)
拉力作用时间段
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114805094586.gif)
由运动学知识
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114805125471.gif)
联立①②可知
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114805172866.gif)
欲使尺子从水平桌面落下必使尺子的重心出桌子边缘。
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114805187398.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114805203771.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241148052031094.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114805234512.gif)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114805250431.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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