题目内容

如图(a)所示,在足够长的光滑水平面上,放置一长为L=1m、质量为m1=0.5kg的木板A,一质量为m2=1kg的小物体B以初速度v0滑上A的上表面的同时对A施加一个水平向右的力FAB之间的动摩擦因数为μ=0.2,g=10m/s2;小物体BA上运动的路程SF力的关系如图(b)所示。求:v0F1F2
(1)4m/s;(2)3N;(3)9N
(1)由图象可看出当F≤1N时,B物体在A板上的路程始终等于板长L,当F=1N时,刚好不从A板右端掉下,此后AB一起相对静止并加速运动。
B物体的加速度为a2A板的加速度为a1,分别由牛顿第二定律:
μm2gm2a2 ············································································································ ①
Fμm2gm1a1 ······································································································· ②
B运动的位移为S2A运动的位移为S1,经过t时间两者速度均为v,根据运动学公式:
SBt ············································································································· ③
SAt ·················································································································· ④
vv0a2ta1t········································································································· ⑤
BA上相对A向右运动的路程SSBSA ······································································ ⑥
联立①②③④⑤⑥解得:S=··············································································· ⑦
F=1N,S=1m代入,解得:v0=4m/s
(2)根据⑦式分析可知,当1N≤FF1时,随着F力增大,S减小,当FF1时,出现S突变,说明此时AB在达到共同速度后,恰好再次发生相对运动,B将会从A板左端掉下。
AB恰好发生相对运动时,B的加速度为a2,则整体加速度也为a2,由牛顿第二定律:
F1=(m1m2)a2········································································································ ⑧
联立①⑧解得解得F1=3N
(3)此时BA上运动的路程为S1==m
FF1时,物体BA板上的路程为B相对A向右运动的路程的两倍。
故当FF2时,将S=0.5S1代入⑦式解得:F2=9N
【评析】本题考查牛顿运动定律。滑块问题是物理模型中非常重要的模型,是学生物理建模能力培养的典型模型。滑块问题的解决非常灵活,针对受力分析、运动分析以及牛顿第二定律的掌握,还有相对运动的分析,特别是摩擦力的变化与转型,都是难点所在。本题通过非常规的图象来分析滑块的运动,能从图中读懂物体的运动。
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