Problem with solving for force constant k of spring?

In summary, to find the force constant of a vertical spring with a 3.60 kg object placed on top, one can use the equation F = mg = -kx, where k is the spring constant and x is the displacement from equilibrium. This method results in a force constant of 1,246.6 N/m. However, if one tries to use the elastic potential energy equation, U = .5kx^2, and solve for k using the work done on the spring, the resulting force constant is 2,493.2 N/m, which is double the first result. This is because energy is not conserved in this system due to the mass oscillating around the equilibrium. Therefore, the first method
  • #1
David Day
12
1
Member advised to use the formatting template for all homework help requests
The question is stated as the following:

When a 3.60 kg object is placed on top of a vertical spring, the spring compresses a distance of 2.83 cm. What is the force constant of the spring?

The correct answer was acquired by using the equation F = mg = -kx, where k is the spring constant and x is the displacement from equillibrium. Using this equation, k is solved as 1,246.6 N/m.

However, in my first attempt at solving, I tried a different method by calculating the amount of work done on the spring and setting that equal to the equation for the elastic potential energy of a spring, U = .5kx2. Calculating the work F*(yf - yi), W = mg*(.0283) = .998 J.
Therefore, .998 = .5kx2. But solving with this method gives k = 2,493.2 N/m, which is not only different from the first result but is exactly double the amount.

What is the reason for this? Why would using this second method not give the correct result for k? Any help/explanation would be greatly appreciated.
 
Physics news on Phys.org
  • #2
Energy is not conserved. In order to come to rest, the energy must dissipate out of the system. Otherwise the mass will oscillate around the equilibrium. What you have computed is the second turning point of such oscillations, which naturally has the spring twice as compressed as the equilibrium.
 
  • Like
Likes David Day
  • #3
The elastic force F is not constant while you are stretching the spring. It is only equal to mg at the very end. Part of ##mg\Delta y## goes into kinetic energy,
 
  • Like
Likes David Day
  • #4
David Day said:
When a 3.60 kg object is placed on top of a vertical spring, the spring compresses a distance of 2.83 cm
That is somewhat open to interpretation. If it is placed on top of the relaxed spring and let go, and the compression distance quoted refers to the maximum distance, then your energy-based method and answer are correct.
But more likely it is intended that the object is only let go gradually, allowing the spring to compress to the equilibrium position without leaving the object any KE.
 
  • Like
Likes David Day

Related to Problem with solving for force constant k of spring?

1. What is the formula for calculating the force constant k of a spring?

The formula for calculating the force constant k of a spring is k = F/x, where F is the applied force and x is the displacement of the spring from its equilibrium position.

2. How do I determine the force constant k if I only know the spring's mass and the period of its oscillation?

If you only know the spring's mass and the period of its oscillation, you can use the formula k = 4π²m/T², where m is the mass of the spring and T is the period of its oscillation.

3. Can the force constant k of a spring change?

Yes, the force constant k of a spring can change depending on factors such as the material of the spring, its length, and its temperature.

4. How does the force constant k affect the behavior of a spring?

The force constant k determines the stiffness of a spring, meaning that a higher force constant results in a stiffer spring that is harder to stretch or compress.

5. What is the unit of measurement for the force constant k?

The unit of measurement for the force constant k is N/m (newtons per meter) in the SI (International System of Units) system.

Similar threads

  • Introductory Physics Homework Help
Replies
14
Views
468
  • Introductory Physics Homework Help
Replies
29
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
442
  • Introductory Physics Homework Help
Replies
14
Views
2K
Replies
5
Views
255
  • Introductory Physics Homework Help
Replies
12
Views
902
  • Introductory Physics Homework Help
Replies
3
Views
449
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
621
Back
Top