Question

If the length of a `42 cm` spring increases to `87 cm` when a `1 kg` weight is hanging from it, what is the spring's constant?

Answer 1

`k=21,778N//m`

Explanation:

Hooke's Law states that the elongation or compression of an elastic spring is directly proportional to the resultant force acting on it and in the direction of the resultant force.
`vecF = k vecx`.

So in this case, the resultant force is the weight force `W=mg=1xx9,8=9,8N`.

The elongation in this case is `x=87-42=45cm`, which is `0,45m` in SI units.

Therefore by Hooke's Law, the spring stiffness constant may be given by

`k=F/x=(9,8N)/(0,45m)=21,778N//m`.

(This is quite a small value, indicating a very elastic spring, easy to stretch and compress. Not very stiff).