Not exactly the best way to describe it but, it is used to calculate resistance of a lever as in the use of a pry bar or pulley. Technology used to increase output with little input.
So what you wanna do is take your two givens. 22 and 3. Now you wanna take 22 and divide that by 3. And that gives you 7.33. Now if your answer HAS to be a whole number it'll be 66.
Complete question:
A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find
(a) the force constant of the spring and (b) the amplitude of the motion.
Answer:
(a) the force constant of the spring = 47 N/m
(b) the amplitude of the motion = 0.292 m
Explanation:
Given;
mass of the spring, m = 200g = 0.2 kg
period of oscillation, T = 0.410 s
total mechanical energy of the spring, E = 2 J
The angular speed is calculated as follows;

(a) the force constant of the spring

(b) the amplitude of the motion
E = ¹/₂kA²
2E = kA²
A² = 2E/k
