Help with spring stiffness calculation (k)

In summary, the problem involves determining the spring stiffness (k) when a steel ball of mass 2kg is released from just being in contact with a compressed spring by 50mm. The given equations are F = mg and Restorative force F = -kx, and the attempted solution used Hooke's Law to find the answer of k = 392.4 N/m. However, the teacher marked it as wrong, suggesting that the inclusion of "initially" implies that the compression given is not the equilibrium position. Conservation of energy could be used to solve the problem instead.
  • #1
imd25
2
0
Hi

I was given the following problem in my coursework:

Homework Statement


"A spring is initially compressed by 50mm when a steel ball of mass 2kg is released from just being in contact with the uncompressed spring. Determine the spring stiffness (k) of the spring."

Homework Equations


F = mg
Restorative force F = -kx

The Attempt at a Solution


I gave the answer below, but the teacher has marked it as wrong. Can anyone shed any light on where I may have made a mistake? I thought it was just a simple application of Hooke's Law.

F = mg
F = 2 x -9.81
F = -19.62 N

F = -kx
19.62 = -k(-0.05)
19.62 = 0.05k
k = 19.62/0.05
k = 392.4

Thanks, any insight gratefully received.
 
Physics news on Phys.org
  • #2
Units?
 
  • #3
Thanks for your reply Bystander,

I did actually include the units (N/m) in my coursework - sorry, I omitted them in my original post by accident.
 
  • #4
I think the inclusion of "initially" implies the compression given is the lowest point the weight reaches rather than the equilibrium position that your solution assumes.
Try applying conservation of energy to solve.
 

What is spring stiffness and why is it important?

Spring stiffness (k) is a measure of the resistance of a spring to deformation when a force is applied. It is important because it determines the amount of force needed to extend or compress a spring, and thus affects the overall performance of a mechanical system.

How do I calculate spring stiffness?

Spring stiffness can be calculated using Hooke's Law, which states that the force applied to a spring is directly proportional to the spring's displacement. The formula for spring stiffness is k = F/x, where k is the spring constant, F is the applied force, and x is the displacement of the spring.

What are the units of spring stiffness?

The units of spring stiffness depend on the units used for force and displacement. In the SI system, the unit for spring stiffness is newtons per meter (N/m). In the US customary system, the unit for spring stiffness is pounds per inch (lb/in).

How does the material of a spring affect its stiffness?

The material of a spring can greatly affect its stiffness. Generally, stiffer materials such as steel will have a higher spring stiffness compared to more flexible materials like rubber. The shape and size of the spring also play a role in determining its stiffness.

Can the stiffness of a spring be changed?

Yes, the stiffness of a spring can be changed by altering its physical properties such as the material, shape, and size. Additionally, the amount of force applied to a spring can also affect its stiffness. Stiffer springs can be achieved by increasing the number of coils or using a thicker wire, while more flexible springs can be achieved by decreasing the number of coils or using a thinner wire.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
368
  • Introductory Physics Homework Help
Replies
2
Views
679
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
449
  • Classical Physics
Replies
6
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
2
Replies
35
Views
2K
Back
Top