Hooke's Law Lab Spring Constant Calculation

Click For Summary
To calculate the spring constant using Hooke's Law, the formula k=F/x is applied, where F is the force (weight) and x is the extension from the equilibrium position. The user is experiencing varying results for the spring constant despite using this formula, leading to confusion about its consistency. It is confirmed that plotting weight against the change in length (Delta x) should yield a straight line, with the slope representing the spring constant. Discrepancies in results may arise from inaccuracies in measuring the initial length of the spring or the weights used. Accurate measurements and consistent methodology are crucial for obtaining a reliable spring constant.
Abhishekkkk__
Messages
2
Reaction score
0

Homework Statement


I need to calculate a spring constant using measurements from a Hooke's Law Apparatus, a spring, and some weights. The weights are hung vertically from the spring and the distance is measured from the equilibrium point of the spring. If I'm solving for k, then k=F/x. I do this for the different lengths of the spring and masses of the weights, and get very different answers for the constant. So here are my questions: 1) Am I even using the right formula? If I'm not, you can go give me the right one. 2) Shouldn't a constant be constant? And help on this matter would be greatly appreciated.

Homework Equations

The Attempt at a Solution


upload_2018-1-27_17-27-24.png
 

Attachments

  • upload_2018-1-27_17-27-24.png
    upload_2018-1-27_17-27-24.png
    18.6 KB · Views: 511
Physics news on Phys.org
Do you get a straight line if you plot weight versus Delta x?
 
Chestermiller said:
Do you get a straight line if you plot weight versus Delta x?

Yes I do
upload_2018-1-27_19-51-22.png
 

Attachments

  • upload_2018-1-27_19-51-22.png
    upload_2018-1-27_19-51-22.png
    3.7 KB · Views: 564
So the spring constant is the slope, and maybe you were off a little on the inextended length.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
7
Views
4K
  • · Replies 14 ·
Replies
14
Views
4K
Replies
9
Views
908
Replies
20
Views
1K
  • · Replies 35 ·
2
Replies
35
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
17
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K