题目内容
如图所示,重为20N的木块放在倾角为θ=300的斜面上受到F =
N的水平恒力的作用做匀速直线运动(F的方向与斜面平行),则木块与斜面的滑动摩擦系数为( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250010371492973.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001037149344.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250010371492973.jpg)
A.0.6 | B.![]() | C.![]() | D.无法确定 |
A
试题分析:对物体受力分析,受拉力、重力、支持力和滑动摩擦力;因为物体做匀速直线运动,故受到平衡力的作用,将重力沿着平行斜面和垂直斜面方向正交分解,垂直斜面方向分力mgcosθ=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001037212572.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250010372271178.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001037259574.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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