Question

If the length of a `27 cm` spring increases to `45 cm` when a `4 kg` weight is hanging from it, what is the spring's constant?

Answer 1

The spring constant is `=217.8kgs^-2`

Explanation:

The equation of the elongation of the spring is

`F=k*Deltax`

The force is `F=4gN`

The elongation is `Deltax=0.45-0.27=0.18m`

The spring constant is

`k=F/(Deltax)=(4g/0.18)Nm^-1`

`=217.8kgs^-2`

as

`1 Nm^-1=1 kgms^-2/m=1 kgs^-2`