A spring with an mm-kg mass and a damping constant 9 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 2.5 newtons. If the spring is stretched 1 meters beyond its natural length and then released with zero velocity, find the mass that would produce critical damping.

4.05 kg

Explanation:

From Hooke's law, force required to stretch the spring is represented as follows:

k (0.5) = 2.5

Spring Constant, k = 5

For critical damping, c² - 4mk = 0

m = c² / 4 k

c = damping constant

m = Mass to produce critical damping

There fore, m = (9²/4*5)

= 81/20

= 4.05 kg

Therefore, the mass that would produce critical damping. = 4.05 kg