an archer pulls her bowstring back 0.400 m by exerting a force that increases uniformly from zero to 230 N. What is the equivalent spring constant of the bow?

i know $F_{SPRING} = kx$ but how does that help me here? F is not constant..

Question

an archer pulls her bowstring back 0.400 m by exerting a force that increases uniformly from zero to 230 N. What is the equivalent spring constant of the bow?

i know $F_{SPRING} = kx$ but how does that help me here? F is not constant..

experimentx · Answer

F = -kx

experimentx · Answer

230 = -k * 0.4

anonymous · Answer

in this case force remains proportional to the extension produced.(Hook's Law)

experimentx · Answer

-230 = -k * 0.4
(i think the left side also be negative)

lgbasallote · Answer

wait wait...what's happening o.O

lgbasallote · Answer

F = -kx? why negative?

anonymous · Answer

Because force is in opposite direction to extension. it is trying to take it back to original position.

lgbasallote · Answer

ohh what i wrote was for compression?

experimentx · Answer

both are same ..

lgbasallote · Answer

anyway...why 230? the fact that it's not costant does not matter?

experimentx · Answer

mechanism is same ... i never calculated hook's constant. from this expression ... it seems that it should be positive http://upload.wikimedia.org/wikipedia/en/math/f/3/3/f337918171cdd650920f50b3b026d4f9.png

lgbasallote · Answer

...so what you're saying is...it should be F = kx?

experimentx · Answer

the force is proportional to displacement from the mean position ... the force will vary constantly.
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