Answer to Question #89442 in Physics for Vera
A spring of natura length 1.5m is extended 0.0050 by a force of 0.8n . What will its length be when the applied force is 3.2n
1
2019-05-09T11:35:40-0400
Hooke's law states
F=kxSo
F2F1=x2x1
x2=F1F2x1
=0.8N3.2N×0.005m=0.02mThe spring length
l2=l0+x2
=1.5m+0.02m=1.52m
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