题目内容
(10分)如图所示,竖直平面内有一粗糙的圆弧
圆管轨道,其半径为R=0.5m,内径很小。平台高h="1.9" m,一质量m=0.5kg、直径略小于圆管内径的小球,从平台边缘的A处水平射出,恰能沿圆管轨道上P点的切线方向进入圆管内,轨道半径OP与竖直线的夹角为3
7°。g=10m/s2,sin37°=0.6,cos37°=0.8。不计空气阻力。求:
(1)小球从平台上的A点射出时的速度v0是多大?
(2)小球通过最高点Q时,圆管轨道对小球向下的压力FQ=3N,小球在圆管轨道中运动时克服阻力所做的功W是多少?![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241218157265469.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412181568072.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412181569572.gif)
(1)小球从平台上的A点射出时的速度v0是多大?
(2)小球通过最高点Q时,圆管轨道对小球向下的压力FQ=3N,小球在圆管轨道中运动时克服阻力所做的功W是多少?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241218157265469.jpg)
(1)v0=8 m/s
(2)W=18.5
J
(2)W=18.5
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412181572672.gif)
(1)设小球从A到P的过程中,运动时
间为t,竖直位移为y,到达P点时竖直方向的速度是vy
,则
y=h-R(1-cos37°) .............................(1分)
.............................(1分)
y=1.8 m,t=0.6 s
vy=gt
.............................(1分)
.............................(1分)
解得v0=8 m/s .............................(1分)
(2)设小球在P点的速度为vP,则
vP=
.............................(1分)
由动能定理有-mgh-W=
..................(2分)
由牛顿第二定律有mg+FQ=
.....................(1分)
vP=10m
/s,vQ=
m/s
解得W=1
8.5 J ........................(1分)
(2)或解:设小球在Q点的速度为vQ,则
L=(h-2R) ....
..........................(1分)
由动能定理有mgL-W=
....................(2分)
由牛顿第二定律有mg+FQ=
.....................(1分)
L=0.9m,vQ=
m/s
解得W=18.5
J .....................(1分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412181569572.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412181577372.gif)
y=h-R(1-cos37°) .............................(1分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121815773433.gif)
y=1.8 m,t=0.6 s
vy=gt
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412181578985.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121815820517.gif)
解得v0=8 m/s .............................(1分)
(2)设小球在P点的速度为vP,则
vP=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121815836401.gif)
由动能定理有-mgh-W=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121815867531.gif)
由牛顿第二定律有mg+FQ=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121815882405.gif)
vP=10m
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412181594565.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121815960209.gif)
解得W=1
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121815976112.gif)
(2)或解:设小球在Q点的速度为vQ,则
L=(h-2R) ....
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412181599265.gif)
由动能定理有mgL-W=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121816007602.gif)
由牛顿第二定律有mg+FQ=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121815882405.gif)
L=0.9m,vQ=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121815960209.gif)
解得W=18.5
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412181605472.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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