Question

If the length of a `24 cm` spring increases to `47 cm` when a `5 kg` weight is hanging from it, what is the spring's constant?

Answer 1

Better to rewrite the question in metres: "`0.24 m` spring increases `0.47 m`..." Spring constant, `k = F/(Delta d) = (m*g)/(Delta d) = (5*9.8)/(0.47-0.24) = 21.3 Nm^-1`

Explanation:

The spring constant, `k`, of a spring is expressed in newton per metre `(Nm^-1)`: for each additional `N` of force the spring expands by `k (m)`.

It's important to use correct SI units for the quantities: centimetre is not a base SI unit, metre is, so convert the `cm` to `m`.

I used `Delta d` to represent the change in length of the spring - final length of `0.47 m` minus initial length `0.24 m`.

The force acting on the spring is the weight force of the mass, which is its mass, `5 kg` times `g = 9.8 Nkg^-1` (more often written as `ms^-2`, but this is an exactly equivalent unit that makes more sense in this context).

Calculating all this leads to a spring constant equal to `21.3 Nm^-1`.