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Hooke's Law, Force Constant Question 
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#1
Mar707, 07:24 PM

P: 66

1. The problem statement, all variables and given/known data
How does the total force constant of two springs hung linearly compare with the individual force constants of springs. Predict the equation that relates the total force constant, ktotal, to the individual force constants, k1 and k2, of two springs joined together linearly. *It's asking this question when a mass is attached to the bottom of one spring and the springs are attached to each other. 2. Relevant equations Fx=kx (Hooke's Law) 3. The attempt at a solution The force constant for two springs hung linearly should be slightly less than the sum of the force constant of the individual springs when they were experimented on without another spring attached. This is because a weight such as 500 g will cause two springs attached together to stretch less than it will cause an individual spring to stretch. The x value for two springs when two springs are attached together will be more than the x value if one individual spring is stretched with a 500 g weight. This is because the force constant for two springs attached together will require more compression or stretch than if an individual spring were to be stretched. 


#2
Mar707, 07:36 PM

Mentor
P: 41,316

Seems like you are all over the place. Think of it this way: The stretch of a spring depends on the tension it is required to exert. If you hang a mass M from one spring, what's the tension? How much does it stretch? Now hang that same mass from two springs strung together. What's the tension in each spring? So how much does each spring stretch? What's the total stretch? Use this line of thought to deduce the spring constant of the double spring. 


#3
Mar707, 09:17 PM

P: 66

The tension is less in each spring so it stretches less. Here's what I got for the spring constant:
2ktotalxtotal = k1x1 + k2x2 When I plugged values from the experiment into this, left side equaled right side. Does that prove the equation? 


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