Question

A 1.37 kg rock is suspended from a spring that is stretched 0.880 m. What is the spring constant for this spring?

Answer 1

By the Hooke's Law, we have:

$$F=kx,$$

here, $F=mg$ is the force that acts on the spring, $k$ is the spring constant, $x$ is the elongation of the sping.

Then, from this formula we can find the spring constant:

$$mg=kx,$$ $$k=\dfrac{mg}{x}=\dfrac{1.37\ kg\cdot 9.8\ \dfrac{m}{s^2}}{0.880\ m}=15.26\ \dfrac{N}{m}.$$

Answer:

$k=15.26\ \dfrac{N}{m}.$