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Force,spring constant,length 
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#1
Aug1909, 12:21 PM

P: 113

1. The problem statement, all variables and given/known data
A 7.50kg ball is placed on top of a spring with a spring constant of 475 N/m that has an initial length of 35.0 cm (h1). When the ball has reached its equilibrium position, it is supported by both springs at a height of 20.0 cm above the table (h3). If the shorter spring has an initial length of 25.0 cm, what is its spring constant? 2. Relevant equations f=kx >attached is the makeshift drawing i made to resemble the actual diagram since it was to big. 3. The attempt at a solution this is what my teacher has told me to do : The magnitude of the force exerted by the second spring is equal to the gravitational force minus the force the first spring exerts on the mass. Divide the force the second spring exerts on the mass by the displacement of that spring when the ball is at equilibrium position to find the spring constant for the second spring. so i have to find the force of both springs whcih i can do for the first one by multiplying 475n/m by 0.35m. how do i find the force of the second one tho? 


#2
Aug1909, 12:27 PM

P: 113

sorry. here is the diagram



#3
Aug1909, 12:37 PM

HW Helper
PF Gold
P: 3,440

You have already posted this same thing under "spring constants"
http://www.physicsforums.com/showthread.php?t=331393 Please follow that thread. 


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