Question

what is the spring constant of a bungee cord that stretches an additional 28m when 300 n of force are applied

Answer 1

The magnitude of the force applied to the spring is equal to (Hook's law)

`F=kx`
where k is the constant of the spring and x is the displacement of the spring with respect to its relaxed position.
In the exercise, the force applied is F=300 N while the displacement is x=28 m. Therefore the spring's constant is

`k= (F)/(x)= (300 N)/(28 m)=10.7 N/m`