Question

A spring of a natural length 3cm is extended by 100cm by a force of 9N,what will be its length when the applied force is10N.

Answer 1

Given:

$l_0=0.03\:\rm m$

$l_1=1.03\:\rm m$

$F_1=9\:\rm N$

$F_2=10\:\rm N$

The Hooke's law says

$$F=k(l-l_0)$$

Hence

$$\frac{F_2}{F_1}=\frac{l_2-l_0}{l_1-l_0}$$

$$l_2=l_0+\frac{F_2}{F_1}(l_1-l_0)\\ =0.03+\frac{10}{9}*1.0=1.14\:\rm m=114\: cm$$