Question

If the length of a `49 cm` spring increases to `73 cm` when a `5 kg` weight is hanging from it, what is the spring's constant?

Answer 1

About `204.17` kilograms per second squared.

Explanation:

We use Hooke's law, which states that,

`F=kx`

where:

• `F` is the force in newtons

• `k` is the spring constant in newtons per meter or kilograms per second squared

• `x` is the extension in meters

So, we get:

`k=F/x`

`=W/x \ (because "the force here is the weight's weight")`

`=(5 \ "kg"*9.8 \ "m/s"^2)/(73 \ "cm"-49 \ "cm")`

`=(49 \ "kg m/s"^2)/(24 \ "cm")`

`=(49 \ "kg m/s"^2)/(0.24 \ "m") \ (because 1 \ "m"=100 \ "cm")`

`~~204.17 \ "kg/s"^2`