Question

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

Answer 1

Hooke's law states

$$F=kx$$

So

$$\frac{F_1}{F_2}=\frac{x_1}{x_2}$$

$$x_2=\frac{F_2}{F_1}x_1$$

$$=\frac{3.2\:\rm{N}}{0.8\:\rm{N}}\times 0.005\:\rm{m}=0.02\:\rm{m}$$

The spring length

$$l_2=l_0+x_2$$

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