A spring with an unknown spring constant is hung vertically, and a 200 g (0.200 kg) mass is attached to the bottom. If the spring stretches 0.250 m from its resting position to the position at which the hanging mass is in equilibrium, what is the spring constant of this spring?

k = 7.84 N/m

Explanation:

We are given;

Mass hanging object; m = 0.2 kg

Extension; Δx = 0.25 m

Now, formula for the force is;

F = k•Δx

Where k is the spring constant

Since we have mass, then F = W = mg = 0.2 × 9.8 = 1.96 N

Thus;

1.96 = k × 0.25

k = 1.96/0.25

k = 7.84 N/m

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!