Illogical conclusion about a mass on a spring

Question

If you place a weight on a spring and it is an equilibrium, then you have this equation:

$mg = kx$

you would solve for k and get:

$k = mg/x$

but, if we used conservation of energy, assuming that when the mass is on the spring it is at x = 0, we would get:

$1/2kx^2 = mgx $ we would get $k = 2mg/x$

What assumption am I making that is false? How would the k value be different with two correct statements?

Welcome to Physics! Please use MathJax to type equations and mathematical symbols. — Níckolas Alves, Dec 12 '22 at 05:44
@josephh Thanks a lot! I realized right when I submitted. That makes everything a lot clearer. Thanks for your help! — Alex Mehta, Dec 12 '22 at 06:02
Does this answer your question? If string is stretched just by weight, where does the gravitational potential energy goes if only half is converted to elastic potential energy? — BowlOfRed, Dec 12 '22 at 06:14
@BowlOfRed I think its a similar idea with the idea that we need to account for kinetic energy, but this was solving for the k value. They are the same concept though. — Alex Mehta, Dec 12 '22 at 06:24

score 1 · Accepted Answer · answered Dec 12 '22 at 05:59

1

You need to account for the KE when the mass is dropped and it moves downward. Unless the spring doesn't move, you can't use this energy equation. And if it doesn't move I don't think you could consider it a linear spring.

answered Dec 12 '22 at 05:59

Alex Mehta

21

Illogical conclusion about a mass on a spring

1 Answers1