Question

If the length of a `20 cm` spring increases to `56 cm` when a `15 kg` weight is hanging from it, what is the spring's constant?

Answer 1

K= 408 N/m

Explanation:

`F= mg = 15 kg xx 9,8 m/(s^2)= 147 N`
For a spring `F= K (X_2-X_1)` so `K= (F/(X_2-X_1))=(147 N)/((56-20) cm)= (147N)/(0,36 m)= 408 N/m`