Finding the spring constant while stretching and compressing

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The discussion revolves around determining the spring constant for a physics lab involving a cart with two springs, one stretched and one compressed. The user has recorded velocities of the cart after applying the springs and is attempting to use the formula k = (1/2mv²)/x to calculate the spring constant. However, they are encountering inconsistent results and are considering using kinetic or potential energy equations instead. Another participant suggests reviewing the basic elastic potential energy equation, Us = 1/2kx², and re-evaluating the approach to ensure the correct equations are applied. The conversation emphasizes the importance of using the right formulas for accurate calculations in physics experiments.
Eric0
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Homework Statement


I have a physics lab that I'm just not making any progress on. For this lab my group used a 314 g cart. Two springs were used, both attached to the cart, and one being stretched the other compressed. The cart was then ran horizontally and it's velocity was found.

I want to find the the spring constant for each trial and then average it. I have the velocities for each trial of the cart after stretching/compressing. The three velocities of the cart after being attached to the stretched or compressed string in m/s are 1.18, 2.68, and 3.65. For the compression trials the velocity in m/s was found to be .2, .7, and 1.0.

Natural length of the stretched spring is 6 cm and the compressed 8.4 cm.

Homework Equations





The Attempt at a Solution



I tried to used the formula given: k= (1/2mv2)/x with x being the distance stretched/compressed. The values I got were kind of funky. I'm thinking I need to use kinetic or potential energy here.
 
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Eric0 said:
I tried to used the formula given: k= (1/2mv2)/x with x being the distance stretched/compressed. The values I got were kind of funky. I'm thinking I need to use kinetic or potential energy here.

Hi Eric,

I think that I might be very wrong here, but I am looking at one of the more basic spring equations here; elastic potential energy: Us = 1/2kx2

…And if it is re-arranged, I cannot find that it turns into one of the equations that you are using here. I might very well be wrong, and I am sure that someone else will be able to help you properly, but I think you need to check you equations and re look at it in a different way.

Liberator.
 
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?

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