题目内容
质量为50 kg的乘客乘坐电梯从一层到四层,电梯自一层启动向上做匀加速运动,加速度的大小是1 m/s2,则电梯启动时地板对乘客的支持力为(g取10 m/s2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241535294553267.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241535294553267.png)
A.600 N | B.550 N |
C.500 N | D.450 N |
B
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241535296736307.png)
试题分析:对乘客受力分析如上图所示,由牛顿第二定律得到
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153529798703.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241535302191329.png)
点评:学生要画出物体的受力示意图,并会在竖直方向分析。
![](http://thumb2018.1010pic.com/images/loading.gif)
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题目内容
A.600 N | B.550 N |
C.500 N | D.450 N |