Question

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?

Answer 1

`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"`