Question

If a 0.6 kg object hanging from a spring stretches it by 0.55 m, then by how much will the spring be stretched if a 1.2 kg object is suspended from it?

Answer 1

`1.1\; {\rm m}` (assuming the mass of the spring is negligible.)

Step-by-step explanation:

Under the assumptions, the force stretching this spring is equal to the weight of the attached object.

With
`g = 9.8\; {\rm N\cdot kg^(-1)}`, the weight of an object would be proportional to its mass. Doubling the mass of the object from
`0.6\; {\rm kg}` to
`1.2\; {\rm kg}` will double its weight.

Assume that this spring satisfies Hooke's Law. The displacement
`x` of the spring from its equilibrium position would be proportional to the force that stretches the spring. Hence, doubling this force will double the displacement of the spring from
`0.55\; {\rm m}` to
`1.1\; {\rm m}`.