2022 英语周报 八年级 GZ 32答案

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(1)对B受力分析可知:km= ng sin,解得:x=m2k(2)A刚挂上挂钩瞬间,B没有获得速度,弹簧弹力不变。但A.B整体瞬间获得加速度,且两者加速度大小相等。对A列牛顿第二定律:2mg-T=2ma对B列牛顿第二定律:T=ma解得:a9(3)物块C刚要离开D时,C与D之间弹力为0,对C受力分析可知,弹簧此时处于拉伸状态,kx= ng sin,解得mg2k则B沿斜面向上移动的距离为d=2+x9kA下落的距离也为d,且A和B速度大小相等,初、末位置弹簧的形变量相等,则弹簧弹性势能相等。根据系统机械能守恒:2mgd- mohsin6=元X 3mv,解得kng答:(1)A没有挂上挂钩前,弹簧的压缩量为2k(2)A刚挂上挂钩瞬间,A的加速度大小为亏g(3)物块C刚要离开D时,B的速度大小为91k°

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