如图所示,绝缘光滑水平轨道AB的B端与处于竖直平面内的四分之一圆弧形粗糙绝缘轨道BC平滑连接,圆弧的半

2025-03-16 13:48:45
推荐回答(1个)
回答1:

(1)设带电体在水平轨道上运动的加速度大小为a,
根据牛顿第二定律有qE=ma
解得 a=
qE
m
=8
m/s 2
设带电体运动到B端的速度大小为v B ,则 v B 2 =2as
解得 v B =
2as
=4m/s

设带电体运动到圆轨道B端时受轨道的支持力为N,根据牛顿第二定律有
N-mg=m
v B 2
R

解得 N=mg+m
v B 2
R
=5N

根据牛顿第三定律可知,带电体对圆弧轨道B端的压力大小N′=N=5N
方向:竖直向下
(2)因电场力做功与路径无关,所以带电体沿圆弧形轨道运动过程中,
电场力所做的功W =qER=0.32J
设带电体沿圆弧形轨道运动过程中摩擦力所做的功为W摩,对此过程根据动能定理有
W + W -mgR=0-
1
2
m v B 2

解得 W =-0.72J
答:(1)带电体运动到圆弧形轨道的B端时对圆弧轨道的压力为5N,方向竖直向下.
(2)带电体沿圆弧形轨道从B端运动到C端的过程中,摩擦力做的功为-0.72J.

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