题目内容
如图所示,绷紧的传送带与水平面的夹角θ=30°,传送带在电动机的带动下,始终保持v0=2 m/s的速率运行.现把一质量为m=1kg的工件(可看为质点)轻轻放在传送带的底端,经时间t=1.9 s,工件被传送到h=1.5 m的高处,并取得了与传送带相同的速度,取g=10 m/s2.求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250027272365650.jpg)
(1)工件与传送带之间的滑动摩擦力f
(2)工件与传送带之间的相对位移Δs
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250027272365650.jpg)
(1)工件与传送带之间的滑动摩擦力f
(2)工件与传送带之间的相对位移Δs
试题分析:(1)由题意得,皮带长为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002727267831.png)
设工件匀加速运动的时间为t1,位移为x1,有
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002727282708.png)
工件做匀速运动,有
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002727282732.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002727298526.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002727314562.png)
所以加速运动阶段的加速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002727345873.png)
在加速运动阶段,根据牛顿第二定律,有
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002727345769.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002727360532.png)
(2)在时间t1内,传送带运动的位移为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002727376683.png)
所以在时间t1内,工作相对传送带的位移为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002727407733.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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