physics question

Question

Two identical springs are connected in parallel. If both springs extend by 0.02m when a 500g mass is added to them, what is the spring constant of each spring?

Answer 1

Solution:

With two springs connected in parallel, the following formula applies:

K_eq = K₁ + K₂ , where K₁ and K₂ are the individual stiffness of each of the springs

K_eq = The equivalent stiffness of both the springs connected in parallel

Knowing that F = K_eq y , where y is the elongation given as 0.02 m

F = mg = K_eq y , (500 g )(kg/1000 g) ( 9.81 m/s²) = K_eq ( 0.02 m) , and

4.91 N = K_eq ( 0.02m ), K_eq = 245.5 N/m,

Knowing that the springs elongate the same distance, then they must have the same stiffness as they share the load, K₁ = K₂ = K

K_eq = K₁+K₂ = 2K , 245.5 N/m = 2 ( K), K = 122.8 N/m