Calculating the Spring Constant and Work Done by an Archer's Bow

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To calculate the spring constant of an archer's bow, Hooke's Law (F = kx) is applied, where k represents the spring constant and F is the maximum force exerted. In this case, the archer pulls the bowstring back 0.400 m with a maximum force of 230 N. The equivalent spring constant can be determined by rearranging the formula to k = F/x. The work done by the archer in pulling the bow is calculated using the formula for spring potential energy, PE = 1/2 kx^2. Understanding these principles allows for accurate calculations of both the spring constant and the work done.
Trista
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An archer pulls her bowstring back .400 m by exerting a force that increases uniformly from zero to 230 N. (a) what is the equivalent spring constant of the bow?
(b) How much work does the archer do in puling the bow?

The spring potential energy is defined by PE = 1/2kx^2. So, what does the k stand for? Force? That doesn't work. x must be the displacement of the string, which is .40, so 1/2k times (.40)^2? Friction isn't an issue, so its non conservative force. I know the theorem, but am having difficulty using it. :cry:

I know work is defined by Force times the displacement(x) and that equals Force cos theta times displacement(x). But I haven't been able to figure out theta.

Thank you, ahead of time, for your help!
Trista
 
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Look at "Hooke's Law." F = kx . k is the spring constant. F (in this equation) is maximum force.
 
Thanks! That was what I was missing, thank you!
 

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