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116问答网 > 半径R=1m的四分之一圆弧轨道下端与一水平轨道连接，水平轨道离地面高度h=0.8m，如图所示，有一质量m=1.0k

半径R=1m的四分之一圆弧轨道下端与一水平轨道连接，水平轨道离地面高度h=0.8m，如图所示，有一质量m=1.0k

2025-04-05 23:59:27

推荐回答（1个）

回答1：

（1）小球B→C
x=v_Bt
h＝

1

2

gt2
得v_B=4 m/s
t=0.4 s
v_y=gt=10×0.4=4 m/s
v_c=

v

+

v

=4

2

m/s
由于水平方向速度与竖直方向速度相等，因此方向与水平夹角45°
（2）根据动能定理，则有mgR-W_f=

1

2

m

v

W_f=1×10×1-12×1×42=2 J
答：（1）不计空气阻力，滑块落在地面上时速度4

2

m/s；方向与水平夹角45°；
（2）滑块在AB轨道上滑行时克服摩擦力做功2 J．

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