AP Physics Summer work problems

AI Thread Summary
To calculate the work performed when drawing a bow modeled as a spring with a spring constant of 168 N/m and a displacement of 54 cm, the correct formula is W = 0.5 * k * x^2, resulting in approximately 24.5 joules. The discussion highlights the importance of using the potential energy equation rather than a simple force equation for this scenario. For the second part of the problem, determining the speed of the 98-gram arrow upon release involves applying the conservation of energy principles. Understanding these concepts is crucial for solving both parts of the problem effectively. Overall, the thread emphasizes the importance of correctly applying physics equations to solve work and energy problems.
Iamaskier721
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Homework Statement



Assuming you have a bow that behaves like a spring with a spring constant of 168 N/m and you pull it to a draw of 54 cm, to the nearest joule how much work do you perform?

Homework Equations



The force of a constant spring is f=-k(x) where x is the displacement and k is the spring constant.

The Attempt at a Solution



So Basically If I find the force and then place it into the simple work equation W=F(x) where x is displacement and F is force. I get 49(rounded). Am I doing this correctly?
 
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There's an equation for potential spring energy... it looks similar to the kinetic energy equation, but with different variables.
 
Welcome to the forums Iammaskier,

Yes, it looks like you are doing it correctly to me. Basically, W = k*x^2 in this case.

There's an equation for potential spring energy... it looks similar to the kinetic energy equation, but with different variables.

Ok, but what does that have to do with this problem?
 
Is use that equation of u=1/2kx^2 and got 24.4944 joules. How does that convert to the work performed?
 
Another quick question! There is a second part which is... to the nearest tenth of a m/s, what is the speed of the 98 gram arrow when it is released?

I know I have to find the velocity but I have no clue how to do it.


I've never taken physics before and it's decenlty hard to teach yourself=/
 
This link shall explain why you simply can't use W=kx^2, but instead W=.5kx^2

http://inventors.about.com/gi/dynamic/offsite.htm?site=http://www.emporia.edu/physics/keithron/collegelab1/hooklaw.htm
 
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And of course, once you realize that, then the second part to your problem should become quite easy if you understand the conservation of energy, as well as a comparison I made in my initial reply regarding another similar looking equation.
 
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