How Do Gravity and Spring Forces Relate in Vertical Spring Energy Calculations?

[gundamn]

Hello! I'm a bit confused as to the relationship between the magnitude of Hooke's law (F=kx) and the magnitude of gravity versus the relationship between gravitational potential energy and spring potential energy for a vertical spring system. So, here is my problem. Imagine that you place a mass on a uncompressed spring. The mass is being held on the weightless platform of the uncompressed spring. Upon releasing the mass and fully compressing the spring, you can calculate the amount of compression of the spring using the fact that the magnitude of the force of gravity will be equal to the magnitude of the spring force pushing back upward:

Fs = Fg
kx = mg
x = mg/k

Now, you can also calculate the energy of the initial and final states of this system. In the initial state, the spring is uncompressed and the mass is motionless, so both kinetic energy and the spring potential energy are zero. Therefore, the system has an initial energy of:

Ei = mgx

where x is the distance the spring will be compressed when the mass is released.

Now, if we calculate the energy of the system in it's final state, when the spring is fully compressed, we know that the potential energy is zero since we used our reference point for zero height as the point where the spring is fully compressed. We also know the kinetic energy is zero because the mass is motionless. This means that the final energy must be:

Ef = (1/2)kx^2

Using conservation of energy, we can say the initial energy of the system equals the final energy of the system. Now this is where my brain explodes. Setting the initial energy equal to the final energy gives:

mgx = (1/2)kx^2
2mg = kx

but i already showed earlier that because of the canceling of the force of gravity by the force of the spring, we know that:

mg = kx

This isn't possible! how can mg = kx AND 2mg = kx? Anyway, it obviously can't, this would mean that half the potential energy magically vanishes once the spring compressed and that can't happen.

Anyway, I know this can't be correct... if anyone can tell what is wrong with this argument, it would bring me great joy :)

Cheers.

 

[tiny-tim]

welcome to pf!

hello gundamn! welcome to pf!

(try using the X² icon just above the Reply box )

i don't see a KE anywhere …

mg = kx is the equilibrium position …

if you use KE + PE_grav + PE_spring = constant, you'll have to add in the KE for the speed the mass is going when it reaches the "zero" position

 

[gundamn]

ooh x², nice :)

I thought that the kinetic energy would be zero in both the initial and final states because just as the mass is released, it is motionless. Also, once the spring compresses as far as it will for the mass, it would also become motionless. So KE = 0 for both the initial and final state..?

 

[tiny-tim]

hi gundamn!

(just got up :zzz: …)

… once the spring compresses as far as it will for the mass, it would also become motionless. So KE = 0 for both the initial and final state..?

but when the spring compresses as far as it will for the mass, it's not in equilibrium, and so kx is not mg …

you're confusing temporary motionlessness with equilibrium ​