Question

A vertically hung spring has a spring constant of 150. newtons per meter. A
2.00-kilogram mass is suspended from the spring and allowed to come to rest.

Calculate the elongation of the spring produced by the suspended 2.00-kilogram mass. [Show all
work, including the equation and substitution with units.]

Answer 1

Based on Hooke's law, the force a spring exerts is negative based on the elongation length. Therefore, since a 2.00 kilogram weight exerts 20 Newtons (gravitation acceleration constant = 10), than the spring must pull back with -20 Newtons. This means that -20 = -kx, where k = spring constant and x = elongation length. k = 150, so that means that x = 20/150, or 2/15 meters.