Hooke's law vs. Elastic potential energy

[**Mariam**]

Homework Statement

I am currently learning about elastic potential energy and this is a question that was given to us by my teacher:

When a 13.2-kg mass is placed on top of a vertical spring, the spring compresses 5.93 cm. Find the force constant of the spring.

Homework Equations

The Attempt at a Solution

To solve this question he used total mechanical energy (ME= GPE+ EPE+ KE)and the conversion of the energy from GPE to EPE (as the object falls down it's energy changes from GPE to EPE and the KE is 0 because the final and initial velocities are zero.)
Because this is a conserved system then ME =0
ΔGPE=-ΔEPE
then he substituted the given into the equations: (13.2)(10) (0-0.0593)=-1/2 (k) (0.0593^2-0^2)
k= 4451.9 N/m

However by using Hooke's law (which I got from the internet- In class we still didn't take hooke's law and all what we know about springs...(elastic) is from EPE equation.): F=kx the force here is Fg=(13.2)(10)=132 132=k(0.0593) k=2225.96 N/m

So my question is why are the answers diffrent what is missing in either of these ways??

 

[brainpushups]

I think the problem is interpreting what is meant by

the spring compresses 5.93 cm

.

I would agree with you that it is more straightforward to use Hooke's law to determine the spring constant in practice. Think about the set up: you place some weights on a spring and slowly lower the spring so that it doesn't bounce around. The spring stretches and you know, by Newton's third law that k x = m g. when the spring is in equilibrium.

What happens if you just add some weight to a spring and then drop it? The spring stretches beyond equilibrium and then begins to oscillate about the (new) equilibrium.

So if you interpret

the spring compresses 5.93 cm

to mean that 5.93 cm is the distance the spring compresses when the weight is dropped then the teacher's approach is valid, though much less practical. I'm sure this was just an exercise in applying conservation laws.