题目内容
某人在地面上用弹簧秤称得其体重为490 N.他将弹簧秤移至电梯内称,0至9s时间段内电梯由静止开始运动,弹簧秤的示数如图所示,求电梯在这9s内的位移。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082415180216325073.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082415180216325073.jpg)
18m, 方向向下
试题分析:由图知在0-3s内人处于失重状态,所以电梯向下做匀加速直线运动,得到
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824151802225718.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241518023341170.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241518024431131.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241518025681188.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824151802989868.png)
在3-6s内人处于正常状态,所以电梯向下做匀速直线运动,得到
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824151803083845.png)
在6-9s内人处于超重状态,所以电梯向下做匀减速直线运动,得到
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824151803192746.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241518033171168.png)
因
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824151803379417.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824151803426657.png)
电梯在这9s内的位移
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241518034891051.png)
点评:学生应根据图像分析出电梯在各时间段的运动性质,再由匀变速直线运动的公式求出相应的物理量。
![](http://thumb2018.1010pic.com/images/loading.gif)
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