题干

如图所示,在天花板上悬挂一轻质弹簧,当在弹簧下端挂上质量m1=0.5kg的重物时,弹簧的长度L1=22cm;当在弹簧下端挂上质量m2=0.75kg的重物时,弹簧的长度L2=23cm.已知弹簧始终在弹性限度内,取g=10m/s2,求该弹簧的劲度系数k.

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答案(点此获取答案解析)

解:设弹簧的自然长度为l0,由胡克定律可得:

F1=k(l1﹣l0)

F2=k(l2﹣l0)

由题意:F1=m1g=0.5×10=5N;F2=m2g

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