题目内容
(12分).如图所示,静止在水平地面上的小黄鸭质量m=20kg,受到与水平面夹角为53°的斜向上的拉力,小黄鸭开始沿水平地面运动。若拉力F=100N,小黄鸭与地面的动摩擦因数为0.2,求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250027234601946.png)
(1)把小黄鸭看作质点,作出其受力示意图;
(2)小黄鸭对地面的压力;
(3)小黄鸭运动的加速度的大小。(sin53°=0.8,cos53°=0.6,g=10m/s2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250027234601946.png)
(1)把小黄鸭看作质点,作出其受力示意图;
(2)小黄鸭对地面的压力;
(3)小黄鸭运动的加速度的大小。(sin53°=0.8,cos53°=0.6,g=10m/s2)
(1)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250027234764594.png)
(2)120N(3)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002723492676.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250027234764594.png)
(2)120N(3)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002723492676.png)
试题分析:(1)如图
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250027234764594.png)
(2)根据平衡条件可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002723523892.png)
所以解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250027235381054.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002723554682.png)
(3)受到的摩擦力为滑动摩擦力,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002723570744.png)
根据牛顿第二定律得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002723585844.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002723492676.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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