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

AI Thread Summary
The discussion focuses on calculating the maximum length of a spring when a 0.5 kg mass is pulled down and given an initial upward speed. The spring constant (k) was determined to be 12.25 N/m using the formula F = -kx. The user initially calculated the maximum extension of the spring to be 1.07 m but found the correct answer to be 1.16 m. A key point raised was the need to incorporate gravitational potential energy into the calculations, which the user was uncertain about. The suggestion was made to set the zero point of gravitational potential energy at the unstretched length of the spring to facilitate the 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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