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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?

Respuesta :

Answer:

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