Calculating Spring Constant with Hook Weight Included

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SUMMARY

The discussion focuses on calculating the spring constant (k) while accounting for the weight of a hook used in a physics experiment. The user measured spring stretch using weights, including a 50-gram hook, leading to inflated k values. The correct approach involves measuring the spring's extension from the hook's original position without weights. The relationship between load (W) and extension (x) is directly proportional, indicating that the hook's weight merely shifts the origin of the graph, not affecting the spring constant itself.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Basic knowledge of force and weight calculations (mg)
  • Familiarity with measurement techniques in physics experiments
  • Ability to manipulate algebraic equations for problem-solving
NEXT STEPS
  • Learn how to accurately measure spring extension from a fixed reference point
  • Study the implications of including additional weights in spring constant calculations
  • Explore graphing techniques for visualizing the relationship between load and extension
  • Investigate methods for correcting experimental data to account for systematic errors
USEFUL FOR

Physics students, educators, and laboratory technicians involved in mechanics experiments and spring constant calculations will benefit from this discussion.

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 height 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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