Question

A spring extends in length by `4.0` `cm` when a mass of `1.5` `kg` is hung from it. What is the spring constant of the spring?

Answer 1

`k=(F_x)/(x-x_0)=(14.7" N")/(0.04" m")=367.5` `Nm^-1`

Explanation:

Let these variables represent the following

`F_x=` the force applied to the spring

`k=` the spring force constant

`x=` the distance from equilibrium

`x_0=` the spring equilibrium position

The equation to determine the spring constant, `k` is

`k=(F_x)/(x-x_0)`

We are given the total measure of elongation is `4` `cm`, which is the difference between `x` and `x_0`. That is,

`x-x_0=4` `cm=0.04` `m`

We are also given the mass of the body, `1.5` `kg`, and assuming this measurement is happening on Earth, we have the acceleration due to gravity `9.8` ms`""^(-2)`.

`F_x=9.8` `(ms^-2)xx1.5` `kg=14.7` `N`

Therefore, the final solution is

`k=(F_x)/(x-x_0)=(14.7" N")/(0.04" m")=367.5` `Nm^-1`