题目内容
质量m=10kg的物体,在F=40N的水平向左的力的作用下,沿水平桌面从静止开始运动,物体运动时受到的滑动摩擦力F’=30N。在开始运动的第6s末撤掉水平力F。求物体从开始运动到最后停止,总共通过的路程。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241303176672864.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241303176672864.png)
24m
解: 由牛顿第二定律得:F- F/=ma…(1分)
m/s2="1" m/s2 (1分)
v=at=1×6=6m/s(2分)
由位移公式
得:(1分)
m (1分)
撤去外力F后, - F/=ma/…(1分)
a/=-3m/s2 (1分)
s (2分)
m (1分)
m (1分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130317698901.png)
v=at=1×6=6m/s(2分)
由位移公式
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130317729771.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130317760762.png)
撤去外力F后, - F/=ma/…(1分)
a/=-3m/s2 (1分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241303177921068.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241303178231152.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130317870823.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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