If the length of a 25 cm spring increases to 67 cm when a 15 kg weight is hanging from it, what is the spring's constant?

1 Answer
Apr 11, 2016

k = 15000/42 = 7500/21 = 2500/7 ("Newtons")/(meters) or
k = 357.14 "Newtons"/"meter"

Explanation:

F_(el) = k*x
F_(el) is the force upon the spring, the "elastic" force;
k is the spring constant;
x is the deformity of the spring;
F_w is the force from the weight;
g is gravity;
m is mass.

Since the length of the spring is 25cm, the deformity is 67 - 25, or 42cm. Using the International Sistem (or S.I.), the deformity must be measured in meters, since the constant(k) is measured in Newtons/meter. So x =0.42 meters.

The force applied to the spring is the weight force from the weight hanging from the spring. So F_w = F_(el). Now, we apply the math:
m*g = k*x
15 * 10 = k * 0.42
k = 150/(42/100)
k = 150/1 * 100/42
k = 15000/42 = 7500/21 = 2500/7 ("Newtons")/(meters) or
k = 357.14 "Newtons"/"meter"