题目内容
(10分)如图所示,质量m=40 kg的木块静止于水平面上,现用恒力F大小为200N,方向与水平方向成q=37°斜向上作用下做匀加速直线运动,2s末撤去拉力木块恰好滑行的距离s
=5.2m,(重力加速度g取10 m/s2,sin37°=0.6,cos37°=0.8.)求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241504074051485.png)
(1) 木块与地面间的动摩擦因数;
(2) 撤去拉力后,木块继续滑行的距离.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150407109226.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241504074051485.png)
(1) 木块与地面间的动摩擦因数;
(2) 撤去拉力后,木块继续滑行的距离.
(1)0.2 (2)6.76 m
试题分析:设木块加速的加速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150407592315.png)
(1)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150407779682.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150407935835.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150408107891.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150408185474.png)
(2)2s末木块的速度:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150408310484.png)
匀减速阶段
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150408450512.png)
木块继续滑行的距离
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150408513645.png)
解得:s=6.76 m (1分)
点评:中等难度。动力学的两类基本问题(1)已知受力情况求运动情况:已知物体的受力情况,由牛顿第二定律求出物体的加速度,再通过运动学公式就可以确定物体的运动情况.(2)已知运动情况求受力情况:已知物体的运动情况,根据 运动学公式求出物体的加速度,再根据牛顿第二定律确定物体所受的力.在处理动力学的两类基本问题时,加速度是联系运动和力的纽带、桥梁.
![](http://thumb2018.1010pic.com/images/loading.gif)
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