问题 问答题

一个平板小车置于光滑水平面上,其右端恰好和一个

1
4
光滑圆弧固定轨道AB的底端等高对接,如图所示.已知小车质量M=3.0kg,长L=2.06m,圆弧轨道半径R=0.8m.现将一质量m=1.0kg的小滑块,由轨道顶端A点无初速释放,滑块滑到B端后冲上小车.滑块与小车上表面间的动摩擦因数μ=0.3  (取g=10m/s2

试求:

(1)滑块到达底端B时,对轨道的压力大小.

(2)通过计算判断滑块是否能滑离小车?若能滑离,求滑离时小车的速度;若不能滑离,求最终滑块相对小车的滑行距离.

答案

(1)滑块从A端下滑到B端,由机械能守恒得

    mgR=

1
2
m
v20

v0=

2gR
=4m/s

在B点,由牛顿第二定律得FN-mg=m

v20
R
    

解得轨道对滑块的支持力FN=3 mg=30 N   

由牛顿第三定律可知,滑块对轨道的压力为30 N 

(2)滑块滑上小车后,假设滑块没有滑出小车二者同速,设速度为v,

由动量守恒:mv0=(M+m)v,得v=1m/s     

由能的转化和守恒得:μmg•△s=

1
2
m
v20
-
1
2
(M+m)v2

滑块在小车上滑行长度△s=2m<L=2.06m

即滑块不能滑离小车

答:(1)滑块到达底端B时,对轨道的压力大小是30N.

(2)滑块不能滑离小车,最终滑块相对小车的滑行距离是2m.

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