Find the elongation of a spring, the stiffness of which is 40 N / m, if a weight of 500 g is suspended from it.

Given:
m = 500 grams = 0.5 kilograms – the mass of the load that was suspended from the spring;
k = 40 Newton / meter – coefficient of spring stiffness;
g = 9.8 meters per second squared – gravitational acceleration (constant near the Earth’s surface).
It is required to determine dx (meter) – the amount of spring elongation.
After the load is suspended from the spring, the spring will elongate under the force of the load’s gravity. This force will be equal to:
F = m * g = 0.5 * 9.8 = 4.9 Newton.
Then, according to Hooke’s law, the elongation of the spring will be:
dx = F / k = 4.9 / 40 = 0.12 meters (the result has been rounded to one hundredth).
Answer: The extension of the spring will be 0.12 meters.