(1)物体与接触面间的动摩擦因数μ=?(2)A、B之间的距离sAB=?

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◎ 题目


(1)物体与接触面间的动摩擦因数μ=?
(2)AB之间的距离sAB=?

◎ 答案

(1)0.58(2)10 m

◎ 解析

(1) 物体在斜面上恰能匀速下滑,平行于斜面方向上的合力为零,则
mgsin30°=μmgcos30°                                                           ①
μ=="0.58··························································" ②
(2) 设物体从AB的运动过程中加速度大小为a1BC的运动过程中加速度大小为a2,则
Fμmgma1··············································· ③
μmgma2··················································· ④
解得a1 m / s2······································ ⑤
a2=10 m / s2
根据题意有····· ⑦
所以········································· ⑧
sAB=10 m···················································· ⑨
评分标准:本题共12分,其中①③④式各2分,②⑤⑥⑦⑧⑨式各1分。

◎ 知识点

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