Calculating Spring Constant with Hook Weight Included

AI Thread Summary
In the physics lab, the spring's stretch was measured using weights attached to a hook, but the hook's weight of 50 grams was not accounted for, leading to inaccurate calculations of the spring constant (k). Measurements taken from the bottom of the hook to the floor resulted in progressively higher values for k as the weight decreased, indicating a potential error in measurement methodology. It is suggested that the extension should be measured from the original position of the spring without any weights, correcting for the hook's weight. The relationship between load and extension is direct, meaning the hook's weight only shifts the graph's origin, not the slope. Accurate determination of the spring constant requires proper measurement of the spring's extension from its initial position.
burnside
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In a physics lab we put weights on the end of a spring and measured how far the spring stretched. We attatched the weight to the spring with a hook that looks more or less like ? and the weights stack on the bottom. To make it easier, i just measured from the bottom of the hook to the floor; so to measure how far the spring stretched, I measured how far the weights were to the floor with the weights on, and then took the weights off, measured from the floor to the bottom of the hook, and assumed the different was how far the spring stretched. Unfortunately the hook weighs 50 grams, which considering we were using weights like 800 grams i assumed it would be negligable.

So, when we measured how far the spring stretched with 870 grams, we really measured how much farther it stretched than when the spring had 50 grams on it.

As the weight becomes closer to zero, the spring constant becomes unreasonably high, i.e.

with 870 grams and 15.1 cm of stretch k=51.9
with 750 grams and 11.5 cm of stretch k=63.9
and with 450 grams and 2.3 cm of stretch k=191

I can't redo my measurements, so I'm wondering if there is any way to solve for what 'k' should really be.

I tried substitution, assuming that the spring stretched an extra 'x' cm for every measurement, i.e.

mg=k(y+x)
.85(9.8)=k(.151+x) .05(9.8)=kx
8.33=(.49/x)(.151+x)
x=.011 meters

but 'k' still gets way too high.

Any help will be appreciated.
 
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burnside said:
To make it easier, i just measured from the bottom of the hook to the floor; so to measure how far the spring stretched, I measured how far the weights were to the floor with the weights on, and then took the weights off, measured from the floor to the bottom of the hook, and assumed the different was how far the spring stretched.
The problem could be that you measured from the weights to the ground and subtracted that from the hight of the hook without the weights. You should have measured from the bottom of the hook both times. That could be the reason that your K got bigger and bigger - because the less weights you stacked the less the difference was. Maybe you could correct that if you knew the length of the weights.
 
It is not necessary to include the additional weight of the hook. Since there is a direct proportional relationship between the load W and the extension x of the spring the effect of the weight of the hook is to offset the origin of your graph just higher up on the line. The extensions need to be measured from the original position of the spring, that is you need to determine the extension from the bottom of the hook when there were no weights on it. If the original height was ho then you need to find the difference of the other heights w.r.t. this ho to get the extensions x as a result of the load of the total weight on the hook at those times.
 
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