Help with spring stiffness calculation (k)

AI Thread Summary
The discussion revolves around calculating the spring stiffness (k) for a spring compressed by 50mm when a 2kg steel ball is released. The initial calculations using Hooke's Law were deemed incorrect, as the assumption made was that the compression distance represented the equilibrium position. A suggestion was made to apply the conservation of energy principle instead of solely relying on Hooke's Law for the solution. The importance of correctly interpreting the problem's wording regarding the point of compression is emphasized. Ultimately, a more accurate approach is recommended to determine the spring stiffness.
imd25
Messages
2
Reaction score
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
Units?
 
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.
 
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.
 
Back
Top