题目内容
如图所示,质量为0.5kg的物体在与水平面成300角的拉力F作用下,沿水平桌面向右做直线运动,经过0.5m的距离速度由0.6m/s变为0.4m/s,已知物体与桌面间的动摩擦因数μ=0.1,求作用力F的大小。(g=10m/s2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241531194961957.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241531194961957.jpg)
0.44N
试题分析:对物体受力分析,建立直角坐标系如图
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241531195111205.png)
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153119574671.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241531196361587.png)
负号表示加速度方向与速度方向相反,即方向向左。
y轴方向
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153119714912.png)
x轴方向 由牛顿第二定律得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153119776878.png)
又
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241531203541136.png)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241531204001141.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241531207591583.png)
点评:本题属于已知运动情况求出加速度,然后根据牛顿第二定律分析受力情况。
![](http://thumb2018.1010pic.com/images/loading.gif)
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