题目内容
已知质量为m的木块在大小为F的水平拉力作用下,沿粗糙水平地面做匀加速直线运动,加速度为a,若在木块上再施加一个与水平拉力F在同一竖直平面内斜向下的推力F,而不改变木块加速度的大小和方向,求(1)木块与地面的动摩擦因数的表达式
(2)设F="30N" , m="2.0kg" , a=5,0m/s2 , 此推力F与水平拉力F的夹角
(2)设F="30N" , m="2.0kg" , a=5,0m/s2 , 此推力F与水平拉力F的夹角
(1)
(2)450
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002236108742.png)
试题分析:第一次物体受力情况如图甲所示,牛顿定律方程为:
F-μfN=ma ① FN=mg ②
由①、②式解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002236108742.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250022361557703.jpg)
第二次物体受力情况如图乙所示,牛顿定律方程为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002236170960.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002236186853.png)
两次a相等,由①、②、③、④式得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250022362022445.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002236217869.png)
故有:θ=arctan
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002236217639.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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