Question

What is the spring constant k if a spring stretches .5m when a 10 kg mass is suspended from it?
F = k x

Answer 1

We can find the spring constant from the Hooke's Law:

$$F=kx,$$ $$mg=kx,$$ $$k=\dfrac{mg}{x}=\dfrac{10\ kg\cdot 9.8\dfrac{m}{s^2}}{0.5\ m}=196\ \dfrac{N}{m}.$$

Answer:

$k=196\ \dfrac{N}{m}.$