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Spring system inside an accelerating box 
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#1
Dec2910, 06:14 PM

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1. The problem statement, all variables and given/known data
A mass m is resting at equilibrium suspended from a vertical spring of natural length L and spring constant K inside a box. The box begins accelerating upward with acceleration a. How much closer does the equilibrium position of the mass move to the bottom of the box? 2. Relevant equations f=kx f=ma 3. The attempt at a solution so this is my f=ma statement kxmg=ma then i solved for x to get x= m(a+g) / k but this answer is wrong and the correct answer is ma / k 


#2
Dec2910, 07:28 PM

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P: 11,678

You are looking for the change in the mass' position from the initial equilibrium position, after the box has begun accelerating. What was its initial position?



#3
Dec2910, 07:32 PM

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In other words, what was the equilibrium position of the mass before the box was accelerated? 


#4
Dec2910, 09:47 PM

P: 7

Spring system inside an accelerating box
hmm the problem doesnt specify that location.
Cant we just assume it to be at location 0 


#5
Dec2910, 10:04 PM

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P: 11,678

Recapping what's happening: 1. Box at rest. Spring unloaded, natural length L. 2. Box at rest. Spring loaded with mass M, settles at equilibrium. 3. Box accelerating. Spring stretches more, assumes new equilibrium. The question is (essentially) asking for the amount of stretching that takes place between items 2 and 3. 


#6
Dec2910, 10:15 PM

P: 7

I think my problem was that i assumed L to be the spring's length when it was loaded. But what you stated makes sense now and it leads to the right answer. excellent help. you gave me info but just enough to make me think and understand it thank you 


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