Spring conservation of energy problem

[mr1709]

1. A spring is suspended from a ceiling and a 256g mass is attached to it and pulled down to stretch the spring by 18.2cm The mass is released and travels through the equilibirum position with a speed of 0.746m/s. Calculate the force of the spring constant

Solution

Et= Et'
Ee = Ek + Eg
0.5kx^2 = 0.5mv^2 + mgy
k = (mv^2 + 2mgy)/x^2

So i was doing this question online and saw the solution and have no issue with solving it after seeing how it was solved. My only issue is regarding the theory of this question and was hoping maybe somoene could explain to me.

So the solution involves equating Ee = EG + Ek, but i don't really understand why eg+ek is equal to Ee. I get that when the spring is stretched down it only has elastic potential energy, but then why when its released does the elastic energy equal the sum of gravitational energy and kinetic energy. And why is the y in mgy equal to the same value as x in the spring formula 0.5kx^2

My only reasoning which i still don't fully understand is that when the spring is in its equlibirium position as its released...Ee= 0 and all the energy is now converted to another form. I initially thought that all the energy became kinetic energy, but why is some gravitational as well?

 

[haruspex]

the solution involves equating Ee = EG + Ek,

It is impossible to evaluate that out of context. What exactly does each of those terms mean here?
A context-free version would be work conservation: ΔEe+ΔEg+ΔEk=0.

Btw, the question as stated appears ambiguous. Is the 18.2cm extension the total extension from relaxed length or additional extension after adding the mass and allowing the system to arrive at equilibrium?

 

[mr1709]

It is impossible to evaluate that out of context. What exactly does each of those terms mean here?

Btw, the question as stated appears ambiguous. Is the 18.2cm extension the total extension from relaxed length or additional extension after adding the mass and allowing the system to arrive at equilibrium?

Its the exact way the question was worded. Il post the websites solution. I may be wrong but i see it as the spring is initially at rest in equlibirum position with the spring already attached. Then someone pulls it down

Solution

Et= Et'
Ee = Ek + Eg
0.5kx^2 = 0.5mv^2 + mgy
k = (mv^2 + 2mgy)/x^2

and once all numbers are plugged in it gives a final spring constant of 31.9. My issue is i don't understand the logic of the problem.

 

[haruspex]

Its the exact way the question was worded. Il post the websites solution.

Solution

Et= Et'
Ee = Ek + Eg
0.5kx^2 = 0.5mv^2 + mgy
k = (mv^2 + 2mgy)/x^2

and once all numbers are plugged in it gives a final spring constant of 31.9. My issue is i don't understand the logic of the problem.

Ok, I see.
Sadly, that is not quite logical.
I'll try taking the given x as the additional extension beyond equilibrium, but in that case the stored PE is not ½kx².
In the equilibrium position, we already have an extension x₀. If the additional extension is Δx then the total stored PE is ½k(x₀+Δx)². So in going from there to equilibrium we have:
ΔEe=½kx₀²-½k(x₀+Δx)²
=-kx₀Δx-½k(Δx)².
This is equal and opposite to the gain in KE and GPE, ½mv²+mgΔx
Since mg=kx₀, cancellation yields
½k(Δx)²=½mv²
So maybe I have taken the wrong meaning for x. Will repost with the other interpretation.

I have no idea why they used x and y for the same entity.

 

[Apashanka]

1. A spring is suspended from a ceiling and a 256g mass is attached to it and pulled down to stretch the spring by 18.2cm The mass is released and travels through the equilibirum position with a speed of 0.746m/s. Calculate the force of the spring constant

If you consider the equilibrium position of the spring to be the zero potential level of the gravitational force ,then at a displacent x from mean position the total energy would be .5kx²_{spring potential}-mgx _{gravitational potential}+KE,
At the time of max. displacement KE=0, and when crossing through mean position PE=0(as x=0),
So .5kx²_max.-mgx_max.=KE_max.(at the mean position)

 

[haruspex]

If you consider the equilibrium position of the spring to be the zero potential level of the gravitational force ,then at a displacent x from mean position the total energy would be .5kx²_{spring potential}-mgx _{gravitational potential}+KE,
At the time of max. displacement KE=0, and when crossing through mean position PE=0(as x=0),
So .5kx²_max.-mgx_max.=KE_max.(at the mean position)

No, ½kx² is the spring potential energy if x is the displacement from the relaxed position, which is different from the equilibrium position.
Recall that in SHM of a suspended mass g does not feature in the period. This is because it cancels out of the ODE. See post #4.

 

[haruspex]

Have now tried the other interpretation of the question, i.e. that the pull down includes the descent from relaxed to equilibrium. I get $(\frac km)^2x^2-(2gx+v^2)\frac km+g^2=0$.
So either way, the solution given on the website is wrong.

 

[Apashanka]

No, ½kx² is the spring potential energy if x is the displacement from the relaxed position, which is different from the equilibrium position.
Recall that in SHM of a suspended mass g does not feature in the period. This is because it cancels out of the ODE. See post #4.

That's right ,
Equilibrium position is at mg/k from the relaxed position.

 

[haruspex]

That's right ,
Equilibrium position is at mg/k from the relaxed position.

So do you agree that what you wrote in post #5 was wrong?

 

[Apashanka]

1. A spring is suspended from a ceiling and a 256g mass is attached to it and pulled down to stretch the spring by 18.2cm The mass is released and travels through the equilibirum position with a speed of 0.746m/s. Calculate the force of the spring constant

The total energy at max. stretch is .5kx_max.²-mgx_max.=E
The total energy at the equilibrium position is .5kx_e²-mgx_e+KE=E ,...(1)
Subtituting x_e=mg/k will give k

 

[haruspex]

The total energy at max. stretch is .5kx_max.²-mgx_max.=E
The total energy at the equilibrium position is .5kx_e²-mgx_e+KE=E ,...(1)
Subtituting x_e=mg/k will give k

Right, but that is not what you wrote in post #5, agreed?
And do you agree that the eqn in post #1 is wrong?

 

[PKM]

k = (mv^2 + 2mgy)/x^2

and once all numbers are plugged in it gives a final spring constant of 31.9.

I think, by equilibrium position the author meant to denote the relaxed position of the spring without the mass. Then the final P.E. of the spring is zero, and for the spring-mass system, $\frac{1}{2}mv^2+mgx+(-\frac{1}{2}kx^2)=0$

But indeed x & y refers to the same length here, if the answer given is 31.9 unit.

 

[Apashanka]

Right, but that is not what you wrote in post #5, agreed?
And do you agree that the eqn in post #1 is wrong?

Yah that's right I have misinterpret it with the simple case of a spring vibrating on a horizontal plane for which the relaxed and the equilibrium position are the same,but it is actually vibration on a vertical plane that's why I have edited it in further posts

 

[haruspex]

I think, by equilibrium position the author meant to denote the relaxed position of the spring without the mass.

Possibly, but that is a quite wrong usage of the term equilibrium position. I feel it is more likely just a blunder in the solution.

 

[PKM]

Possibly, but that is a quite wrong usage of the term equilibrium position. I feel it is more likely just a blunder in the solution.

Yes, I feel the same $\ddot\smile$