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1.(2012郑州市预测卷)如图所示,一根轻绳跨过光滑定滑轮,两端分别系一个质量为m1、m2的物块m1放在地面上,m2离地面有一定高度当m2的质量发生改变时,m1的加速度a的大小也将随之改变以下的四个图象,最能准确反映a与m2间的关系的是() 解析:选D.当m2m1时,系统静止,a0,当m2m1时,对整体有m2gm1g(m1m2)a即ageq f(2m1,m1m2)g,可见m2时 ,ag,故D正确2起跳摸高是学生常进行的一项活动小亮同学身高1.72 m,体重60 kg,站立时举手达到2.14 m他弯曲两腿,再用力蹬地,经0.4 s竖直跳起,设他蹬地的力大小恒为1050 N,不计空气阻力,取g10 m/s2,求小亮同学起跳摸高的最大高度是多少?解析:小亮同学起跳摸高包含两个过程:第一阶段用力蹬地获得一定的初速度,第二阶段竖直上抛达最大高度蹬地时,由Fma知Fmgma1a17.5 m/s2v1at3.0 m/s竖直上抛heq f(voal(2,1),2g)0.45 m所以摸高Hh0h2.59 m.答案:2.59 m3如图甲所示,用一水平外力F拉着一个静止在倾角为的光滑斜面上的物体,逐渐增大F,物体做变加速运动,其加速度a随外力F变化的图象如图乙所示,若重力加速度g取10 m/s2.根据图乙中所提供的信息计算(1)物体的质量m;(2)斜面的倾角.解析:对物体受力分析,根据牛顿第二定律得:aeq f(F,m)cosgsin,当F0 N时a6 m/s2,当F20 N时a2 m/s2,解得37,m2 kg.答案:(1)2 kg(2)374如图所示,薄木板A长l5.0 m,质量M5.0 kg,放在水平桌面上,板右端与桌面相齐,在A上距右端s3.0 m处放一小物块B,质量m2.0 kg,已知A、B间动摩擦因数10.1,A与桌面间和B与桌面间的动摩擦因数20.2,原来系统静止,现对平板A施加一个水平向右大小恒定的拉力F,作用一段时间,将平板A从小物块B的下面抽出,且小物块B最后恰好停在桌面的右端边缘取g10 m/s2,求:(1)B运动到桌边的时间t;(2)拉力F的大小解析:(1)对于B,在未离开A时,其运动加速度大小为:a1eq f(1mg,m)1g1 m/s2B从A上掉下后,B在桌面的摩擦力作用下做减速运动,其加速度大小为a2eq f(2mg,m)2g2 m/s2设B从A上掉下时,其速度为v,则由题给条件应有:eq f(v2,2a1)eq f(v2,2a2)s代入s及a1、a2的值解得:v2 m/s设自A开始运动至B从A上掉下的这一过程中,经历的时间t1,B减速运动到桌边经历时间为t2,则t1eq f(v,a1)2 s,t2eq f(v,a2)1 s故B运动的时间是:tt1t23 s. (2)设自A开始运动至B从A上掉下的这一过程中,B的位移为s1B.A的位移为s1A,则:s1Beq f(v2,2a1)2 ms1Alss1B2 m2 m4 m以aA表示这段时间内A的加速度,则有s1Aeq f(1,2)aAteq oal(2,1)对A由牛顿第二定律应有F1mg2(Mm)gMaA代入数据解得F26 N.答案:(1)3 s(2)26 N5(2011南平检测题)广州亚运会开幕式文艺表演最震撼人心的节目当数白云之帆篇章,是由地面上近1400人用绳索拉着的180名演员上演着“空中飞人”九分钟的时间里,180名塔沟武校的武林小子在帆屏上时而俯冲而下,时而攀沿升空,演绎了一场世所罕见的惊险、刺激却又浪漫、温馨的节目现在把他们某次训练过程中的情节简化成如下模型:如图所示,地面上的人通过定滑轮用轻绳将质量m 60 kg的演员从静止竖直向上拉高24 m,用时t6 s,演员在竖直方向上先做匀加速直线运动,再做匀减速直线运动后静止在高空其加速和减速过程中的加速度之比为15,忽略滑轮的质量和滑轮与轴之间的摩擦力求(1)演员在匀加速上升过程中的加速度(2)匀加速上升过程中绳子对地面上的人的拉力为多少?(g10 m/s2)解析:设演员在匀加速和匀减速过程中所用时间分别是t1和t2,加速度大小分别为a和5a,因为匀加速的末速度即为匀减速的初速度,所以有:(1)at15at2又因为t1t26 s所以可得:t15 st21 sheq f(1,2)ateq oal(2,1)eq f(1,2
B.5/21/2
C.2
D.21/2
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