Question

A spring of natural length 3m is extended by 0.01m by a force of 4N what will be it's length when the applied force is 12N

Answer 1

Let's first find the spring constant from the Hooke's law:

$$F=kx,$$ $$k=\dfrac{F}{x}=\dfrac{4\ N}{0.01\ m}=400\ \dfrac{N}{m}.$$

Then, we can find the elongation of the spring when the applied force is 12 N:

$$x=\dfrac{F}{k}=\dfrac{12\ N}{400\ m}=0.03\ m.$$

Therefore, the length of the spring will be:

$$l=x_0+x=3\ m+0.03\ m=3.03\ m.$$