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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

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leena

Hi there!

Using Hooke's Law:
[tex]F = kx[/tex]

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

We are given the force and displacement, so solve for 'k':
[tex]k = \frac{F}{x}\\ \\ k = \frac{230}{0.4} = \boxed{575 N/m}[/tex]

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