题目内容
)如图所示,质量为m=4kg的物体与水平地面间的动摩擦因数μ=0.2,现用F=25N与水平方向成θ=370的力拉物体,使物体由静止开始做匀加速直线运动。求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241152430081582.jpg)
(1)物体所受支持力为多大?摩擦力为多大?
(2)求物体的加速度的大小?
(3)若F作用t=4s后即撤除,此后物体还能运动多久?
(sin37°=0.6,cos37°=0.8,g="10" m/s2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241152430081582.jpg)
(1)物体所受支持力为多大?摩擦力为多大?
(2)求物体的加速度的大小?
(3)若F作用t=4s后即撤除,此后物体还能运动多久?
(sin37°=0.6,cos37°=0.8,g="10" m/s2)
(1)FN="25N " , f=5N (2)a=3.75m/s2(3)7.5
对物体进行受力分析,建立坐标系如图,分别沿水平,竖直方向分解力 F。(1分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241152430241270.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115243039450.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115243055582.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115243071597.gif)
有(3分) ; ;代入数据可得:(1)FN="25N " , f="5N " (2分)
(2)a=3.75m/s2(1分)
(3)F作用t=4s时的速度v1=at=3.75×4m/s=15m/s(1分)
撤除后受力分析,受重力G,支持力FN1和摩擦力f1的作用,如图所示。(1分)
f1=μFN1=μmg(1分)
(1分)
设此后物体还能运动t1s
由v1=at1解得t1=7.5s(1分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241152430241270.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115243039450.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115243055582.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115243071597.gif)
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(2)a=3.75m/s2(1分)
(3)F作用t=4s时的速度v1=at=3.75×4m/s=15m/s(1分)
撤除后受力分析,受重力G,支持力FN1和摩擦力f1的作用,如图所示。(1分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115243086680.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115243102711.gif)
设此后物体还能运动t1s
由v1=at1解得t1=7.5s(1分)
![](http://thumb2018.1010pic.com/images/loading.gif)
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