一道物理高一题目
\u8fd9\u662f\u4e00\u9053\u9ad8\u4e00\u7684\u7269\u7406\u9898\uff1a\u8bbe\u52a0\u901f\u9636\u6bb5\u7528\u65f6\u4e3at
\u5219\u52a0\u901f\u9636\u6bb5\u7ecf\u8fc7\u8def\u7a0b\u4e3as=\uff081/2\uff09gt² \u6700\u5927\u901f\u5ea6\u4e3av=gt
\u51cf\u901f\u9636\u6bb5\u52a0\u901f\u5ea6a=\uff08975N-650N\uff09/65kg=5m/s²
\u51cf\u901f\u9636\u6bb5\u7528\u65f6t`=\uff08v-6m/s\uff09/\uff085m/s²\uff09
\u51cf\u901f\u9636\u6bb5\u8def\u7a0bs`=vt`+\uff081/2)at`² s+s`=11.4
\u4ee3\u5165\u6570\u636e\u53ef\u7b97\u51fa\uff1at=3.6s
v=gt=36m/s t\u603b=t+t`=3.6+6=9.6s
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再对A分析，设弹簧弹力为F，
那么F1-F=ma
得到F=15N
因此弹簧伸长量x=F/k=0.05m=5cm
所以AB之间的距离为S=15cm+5cm=20cm
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