题目内容
如图所示,水平桌面上的物体A,质量为m1,与桌面的滑动摩擦系数为μ,用细绳跨过定滑轮与质量为m2的物体B连接,设B的加速度为a.把B移去,用一竖直向下的拉力F=m2g代替B,设这时A的加速度为a′,则:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241437356563078.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241437356563078.png)
A.a>a′ | B.a′>a; |
C.a=a′ | D.无法确定. |
B
由牛顿第二定律
,当撤去物体B后
比较可知B对;
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824143735719971.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824143735765901.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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题目内容
A.a>a′ | B.a′>a; |
C.a=a′ | D.无法确定. |