Question

The spring under the action of a force of 4 N lengthened by 5 mm. Determine the weight of the weight that will extend this spring by 16 mm.

Answer 1

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

$$F=kx,$$ $$k=\dfrac{F}{x}=\dfrac{4\ N}{5\cdot10^{-3}\ m}=800\ \dfrac{N}{m}.$$

Finally, we can find the weight of the weight that will extend this spring by 16 mm:

$$F=kx=800\ \dfrac{N}{m}\cdot16\cdot10^{-3}\ m=12.8\ N.$$