Question

A spring (k = 3.5N/m) is attached to a superman doll of mass = 13.0 kg. If the action figure is hung from the ceiling by this spring, how much would the spring be stretched?

Answer 1

When the doll is in equilibrium, the force tension in the spring (by Hooke's law) is equal to the weight:

$$kx=mg.$$

The extension of the spring is

$$x=\frac{mg}k=\frac{13·9.8}{3.5}=36.4\text{ m}.$$