Answer to Question #89442 in Physics for Vera

Question #89442
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
Expert's answer
2019-05-09T11:35:40-0400

Hooke's law states

F=kxF=kx

So

F1F2=x1x2\frac{F_1}{F_2}=\frac{x_1}{x_2}

x2=F2F1x1x_2=\frac{F_2}{F_1}x_1

=3.2N0.8N×0.005m=0.02m=\frac{3.2\:\rm{N}}{0.8\:\rm{N}}\times 0.005\:\rm{m}=0.02\:\rm{m}

The spring length

l2=l0+x2l_2=l_0+x_2

=1.5m+0.02m=1.52m=1.5\:\rm{m}+0.02\:\rm{m}=1.52\:\rm{m}


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