How Is the Maximum Length of a Spring Calculated with a Hanging Mass?

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Homework Help Overview

The problem involves a mass hanging from a vertical spring, where the initial conditions include the mass being pulled down and given an initial speed. The goal is to determine the maximum length of the spring during the subsequent motion, with specific values provided for mass, spring length, and speed.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the spring constant and the application of energy conservation principles, including kinetic and potential energy. There is uncertainty regarding the inclusion of gravitational potential energy in the calculations, with some participants questioning how to incorporate it without knowing the final length of the spring.

Discussion Status

Participants are actively engaging with the problem, exploring different aspects of energy conservation and questioning assumptions related to gravitational potential energy. Guidance has been offered regarding expressing potential energy in terms of the final length, indicating a productive direction in the discussion.

Contextual Notes

There is an ongoing discussion about the correct approach to account for gravitational potential energy, with participants noting the need to establish a reference point for potential energy calculations.

Jacob959
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A mass of 0.5 kg hangs motionless from a vertical spring whose length is 0.80 m and whose unstretched length is 0.40 m. Next the mass is pulled down to where the spring has a length of 1.00 m and given an initial speed upwards of 1.5 m/s. What is the maximum length of the spring during the motion that follows?



2. Uspring = .5kx^2, KE = .5mv^2




3.First I found the k value of the spring by taking F=-kx, or mg=kx and got 4.9=k(.4) or k = 12.25. Then, I thought the KEinitial + Uinitial = Ufinal. So I solved the equation .5(12.25)(.6)^2+.5(.5)(1.5)^2 = .5(12.25)(x)^2 to find x, or the final chance in the length of the spring. Once I found this, which I got to be.672, I added this to the unstretched length to get 1.07 as the max length of the spring. However, the correct answer is 1.16 and I can't figure out what I'm doing wrong. Any help?
 
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Jacob959 said:
A mass of 0.5 kg hangs motionless from a vertical spring whose length is 0.80 m and whose unstretched length is 0.40 m. Next the mass is pulled down to where the spring has a length of 1.00 m and given an initial speed upwards of 1.5 m/s. What is the maximum length of the spring during the motion that follows?



2. Uspring = .5kx^2, KE = .5mv^2




3.First I found the k value of the spring by taking F=-kx, or mg=kx and got 4.9=k(.4) or k = 12.25. Then, I thought the KEinitial + Uinitial = Ufinal. So I solved the equation .5(12.25)(.6)^2+.5(.5)(1.5)^2 = .5(12.25)(x)^2 to find x, or the final chance in the length of the spring. Once I found this, which I got to be.672, I added this to the unstretched length to get 1.07 as the max length of the spring. However, the correct answer is 1.16 and I can't figure out what I'm doing wrong. Any help?

Do not forget the change of gravitational potential energy.

ehild
 
I thought of that, but I am entirely uncertain how to add that in because we won't know the potential of gravity unless we know the final length, right? How would I add this in?
 
Express it in terms of x.

ehild
 
You can place the zero of gravitational potential energy at the end of the unstretched spring. Initially, the potential energy is -0.6mg, and the final potential energy is -mgx.

ehild
 

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