Elastic Potential Energy problem

AI Thread Summary
To determine how far a spring with a force constant of 30000 N/m must be stretched to achieve a potential energy of 47 J, the equation Us = 1/2 k(∆x)^2 is used. The calculation initially led to an incorrect step where subtraction was applied instead of division. The correct approach involves dividing 47 by 15000, resulting in approximately 0.0031, which should then be squared to find the stretch distance. The final value was clarified to be around 0.0557 m after correcting earlier mistakes. Properly applying the formula and ensuring correct arithmetic is crucial for accurate results.
kadavis2010
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Homework Statement


A spring has a force constant of 30000 N/m. How far must it be stretched for it's potential energy to be 47 J? Answer in units of m.

Homework Equations


Us= 1/2k(∆x)^2

The Attempt at a Solution


47= 1/2 (30000)(x)^2
47=15000(x)^2
-14953= x^2
√14953= 122.28
 
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kadavis2010 said:

The Attempt at a Solution


47= 1/2 (30000)(x)^2
47=15000(x)^2
So far, so good.
-14953= x^2
Oops... you subtracted instead of divided. Redo this step.
 
47= 1/2 (30000)(x)^2
47=15000(x)^2
-14953= x^2
√14953= 122.28
You subtracted when you should have divided.
 
So you are saying to do 47/15000?
that would equal .0031... then would i do the square root of it or leave it at that?

ohhh nvm i did that on one of my attempts but i forgot to put .0 in front of .0557=] Thanks a lot. I apperciate it!
 

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