An archer pulls her bow string back 0.390 m

Question

An archer pulls her bow string back 0.390 m by exerting a force that increases uniformly from zero to 241 N. What is the equivalent spring constant of the bow?

How much work is done in pulling the bow?

Answer 1

The spring force F (-) kx, 241 N = (0.390 m) k, k = 241/0.390 = 618 N/m.

The work done is the change in Elastic Potential Energy of the system, W = 1/2 k x² - 0 = 1/2 (618 N/m) (0.390 m)² = 47.0 J.