Question

A 1.55 kg rock is suspended from a spring that is stretched 0.870 m. What is the spring constant for this spring?

Answer 1

By definition, the spring constant is:

$$k=\dfrac{P} {x}$$

where $P=mg$ is the weight of the rock, $x=0.87m$ is the elongation of the string, $m=1.55kg$ is the mass of the rock, and $g=9.81m/s^2$ is the gravitational acceleration. Thus, obtain :

$$k=\dfrac{mg} {x} \\ k=\dfrac{1.55\cdot9.81} {0.87}\approx17.48N/m$$

Answer. 17.48 N/m.