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 ​

 

[rwisz]

Hello there! It seems like you are experiencing some confusion regarding the relationship between Hooke's law, gravity, and potential energy in a vertical spring system. Let me try to clarify things for you.

First of all, it is important to understand that Hooke's law (F=kx) and the force of gravity (Fg=mg) are two separate concepts and should not be equated. Hooke's law describes the relationship between the force exerted by a spring and the displacement of the spring from its equilibrium position. On the other hand, the force of gravity is the force exerted by the Earth on an object due to its mass.

In the scenario you described, the force of gravity and the spring force are in equilibrium, meaning they are equal in magnitude but opposite in direction. This is why you can set Fs (spring force) equal to Fg (force of gravity). However, this does not mean that kx is equal to mg. They are two different quantities and should not be equated.

Moving on to the concept of potential energy, it is important to note that potential energy is a relative quantity and depends on the choice of reference point. In your initial state, you have chosen the uncompressed spring as your reference point, which means the potential energy is zero. However, in your final state, you have chosen the fully compressed spring as your reference point, which means the potential energy is not zero. This is why you cannot set the initial energy (mgx) equal to the final energy ((1/2)kx^2) since they are calculated at different reference points.

In conclusion, there is nothing wrong with your argument. However, you need to be careful with equating different quantities and choosing your reference points when calculating potential energy. I hope this helps clarify things for you. Keep exploring and questioning, that's what science is all about!