Not sure why my method doesn't work: Springs, Potential Energy and Work

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SUMMARY

The discussion centers on the calculation of elastic potential energy (EPE) using springs, specifically addressing the discrepancy in results when applying different methods. The user initially attempted to calculate work done using the formula Ee = xF, where x represents the distance and F is the force, resulting in an incorrect value of 2.7J. The correct approach involves recognizing that the force exerted by a spring is not constant and requires integration over the distance compressed, particularly when the final force is 18 N at a compression of 15 cm.

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ericcy
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Homework Statement
A force of 18N compresses a spring by 15cm. By how much does the spring's potential energy change?
Relevant Equations
Ee=1/2kx^2, Fs=kx
I know that you can get the answer through using Fs as 18 and solving for K, then subbing it into the equation for elastic energy. I was just wondering why another method wouldn't work.

I tried doing it using the concept that Work is an equal to the Change in Elastic Energy, therefore Ee=xF, because x should be equal to the distance it traveled. I then got the answer of 2.7J which is wrong, I just don't know why that way wouldn't work.

All responses are appreciated, unit test tomorrow.
 
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ericcy said:
I tried doing it using the concept that Work is an equal to the Change in Elastic Energy, therefore Ee=xF, because x should be equal to the distance it traveled. I then got the answer of 2.7J which is wrong, I just don't know why that way wouldn't work.
Presumably the 18 N force is the final force holding the spring compressed to 15 cm. While being compressed, the force will change from 0 N to 18 N. Since the force is not constant, you can't apply a simply W = Fd approach. You would need to integrate over the trajectory.
 
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gneill said:
In future, when you set your thread titles please make sure that they are descriptive of the specific problem or at least area of physics involved. In this particular case you might have mentioned springs, potential energy and work.
Thread title fixed up. :smile:
 
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gneill said:
Presumably the 18 N force is the final force holding the spring compressed to 15 cm.
Yes, but of course that is just a presumption. E.g. it could be that a weight is already sitting on top of the spring when the 18N is applied, in which case the gain in spring PE is greater. Or if it starts with a mass hanging from the spring and the 18N is applied upwards then the EPE could decrease (the 15cm 'compression' being actually a reduction in extension)!
A poorly worded question: it could lead some students to think the EPE change can always be determined directly from the changes in load and compression.
 
haruspex said:
Yes, but of course that is just a presumption. E.g. it could be that a weight is already sitting on top of the spring when the 18N is applied, in which case the gain in spring PE is greater. Or if it starts with a mass hanging from the spring and the 18N is applied upwards then the EPE could decrease (the 15cm 'compression' being actually a reduction in extension)!
A poorly worded question: it could lead some students to think the EPE change can always be determined directly from the changes in load and compression.
Yes, I agree with that assessment. It's sometimes difficult to determine what level of understanding the question is aimed at, or what unstated initial conditions apply. I assumed a basic introductory physics situation, since this is the introductory physics forum. But I've been tripped up in this regard in the past. Eh. I try to interpret the given problem in the light of the given statement and the "level" of the forum it's posted in.

Poorly worded or inadequately bounded questions seem to be (unfortunately) frequent for high school or first year university physics programs.
 

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