Question

An archer pulls her bow string back 0.40 m by exerting a force that increases uniformly from zero to 230 N. (a) What is the equivalent spring constant of the bow

Answer 1

Hi there!

Using Hooke's Law:

`F = kx`

F = Force (N)
k = Spring constant (N/m)
x = displacement from equilibrium

We are given the force and displacement, so solve for 'k':

`k = (F)/(x)\\ \\ k = (230)/(0.4) = \boxed{575 N/m}`