Hooke's Law, Force Constant Question

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SUMMARY

The discussion focuses on the relationship between the total force constant (ktotal) of two springs arranged linearly and their individual force constants (k1 and k2) as described by Hooke's Law (Fx = kx). It is established that when two springs are connected in series, the effective spring constant is less than the sum of the individual spring constants. The equation derived from the analysis is 2ktotalxtotal = k1x1 + k2x2, demonstrating that the tension in each spring is the same, but the total stretch is the sum of the stretches of each spring.

PREREQUISITES
  • Understanding of Hooke's Law (Fx = kx)
  • Knowledge of spring constants (k1, k2)
  • Basic principles of tension in mechanical systems
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Research the concept of springs in series and their effective spring constants
  • Learn about the implications of Hooke's Law in real-world applications
  • Explore the mathematical derivation of spring constants in series and parallel configurations
  • Investigate experimental methods for measuring spring constants
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Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in understanding the behavior of springs in mechanical systems.

BayernBlues
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Homework Statement



-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.

Homework Equations



Fx=kx (Hooke's Law)

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.
 
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BayernBlues said:
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.
Slightly less?

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.
Do they stretch less? Or more?

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.
So they stretch more?

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.
 
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?
 
BayernBlues said:
The tension is less in each spring so it stretches less.
The tension is the same in each spring. But the total stretch of the two-spring system will be the sum of the stretches of each individual spring. Use that reasoning to find the effective spring constant of the system
 

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