Question

In preparing to shoot an arrow, an archer pulls a bow string back 0.424 m by exerting a force that increases uniformly from 0 to 223 N. What is the equivalent spring constant of the bow?

Answer 1

Answer:

525.94N/m

Explanation:

According to Hooke's law, the extension or compression of an elastic material is proportional to an applied force provided that the elastic limit of the material is not exceeded.

$F=ke..................(1)$

where  F is the applied force or load, k is the elastic constant or stiffness of the material and e is the extension.

In this problem, the bow string is assumed to behave like an elastic material that is stretched not beyond the elastic limit.

Given;

F = 223N;

e = 0.424m

k = ?

We make substitutions into equation (1) and then solve for k.

$223=k*0.424\\k=\frac{223N}{0.424m}\\k = 525.94N/m$