Springs - calculating K constant

[iGoRaw]

I'm so confused...
I've done so many spring questions i've started to lose understanding now.
Someone please clarify.. When a spring is hanging vertically and is stretched.. let's say from its natural length of 40 cm to 70 cm by adding a mass of 1kg. Why can't i relate the gravitational potential energy which is mg*deltaH = 1*10*0.3 = 3 J, to the springs Strain potential energy, 0.5Kx^2 = 0.5k*0.3^2 ?? If you do this you get a K constant of 66.67 N m^-1.
But of course that's the wrong answer, otherwise i wouldn't be asking. What the answers do is relate the graviational potential energy (EGP=3 J) to Fx, As W=EGP, Therefore 3=F*0.3, F=10 N. Then they relate F=Kx, hence 10=K*0.3, and K=33.33 N m^-1.
I see that if i divide my answer of 66.67 by 2 i get 33.33, but why do i need to do this? Am i doing something wrong?

[SocialRhubarb]

The problem is that you've assumed that gravitational potential is only converted to elastic potential in the spring.

Imagine that you have an object on a spring, which you're holding, and the spring is at its natural length. Then, lets say you let go of the object. What happens to the object? It begins to oscillate up and down.

Why is this relevant? Because when the object falls down it doesn't only have elastic potential energy, but it has kinetic energy as well. Think about what this means at the point where the weight force down is equal to the force of the spring pulling up. First of all, you've stretched the spring somewhat, since it's exerting a force, so we do have energy stored in the spring. But you've also got kinetic energy, since when the weight force is equal to the force of the spring, it's velocity isn't 0, its acceleration is 0. The object would be accelerating downwards until it reached that point, actually be at its maximum speed when it is at that point.

So the change in gravitational potential isn't equal to the energy stored in the spring, because it would actually have kinetic energy at that point too. Hence, it is much simpler if we just use F=kx, as you've recognised, to solve this question.

[iGoRaw]

I fully understand that.. GRRRRRRR this is why im so confused. It's because i understand that.
Here is the problem.. In the attached picture, i tried using the Fx alternative to calculate the K constant. But instead they RELATE SPE AND GPE!!!! They do the following... EGP=18*70*10=12600=0.5K*8^2, therefore K=394.
Help???

[lzxnl]

I'm so confused...
I've done so many spring questions i've started to lose understanding now.
Someone please clarify.. When a spring is hanging vertically and is stretched.. let's say from its natural length of 40 cm to 70 cm by adding a mass of 1kg. Why can't i relate the gravitational potential energy which is mg*deltaH = 1*10*0.3 = 3 J, to the springs Strain potential energy, 0.5Kx^2 = 0.5k*0.3^2 ?? If you do this you get a K constant of 66.67 N m^-1.
But of course that's the wrong answer, otherwise i wouldn't be asking. What the answers do is relate the graviational potential energy (EGP=3 J) to Fx, As W=EGP, Therefore 3=F*0.3, F=10 N. Then they relate F=Kx, hence 10=K*0.3, and K=33.33 N m^-1.
I see that if i divide my answer of 66.67 by 2 i get 33.33, but why do i need to do this? Am i doing something wrong?

What's the wording of the original question (first one)?
The problem with your interpretation is that if you just add a mass of one kilogram, you can't use conservation of energy as the spring won't stay still. The question sounds like the spring remains still at the end, which implies external energy was required to keep it still. If the spring remains still, you resolve forces.
If the spring doesn't remain still, you have an entirely different question.

I fully understand that.. GRRRRRRR this is why im so confused. It's because i understand that.
Here is the problem.. In the attached picture, i tried using the Fx alternative to calculate the K constant. But instead they RELATE SPE AND GPE!!!! They do the following... EGP=18*70*10=12600=0.5K*8^2, therefore K=394.
Help???

This time, it's because forces don't help. Think about it this way. When going bungee jumping, at the very bottom, the net force is NOT zero. You may be stationary, but you're accelerating upwards. That's why forces won't help you there. If you know about simple harmonic motion, you'll see that actually, the net force IS zero right in the middle of your motion, but that's not necessary here.

For this question, you're given an extension for the spring and a difference in height. The person is still at the beginning and at the very bottom. This is why we only consider potential energies as no energy goes into kinetic.
As the person falls, they lose gravitational potential energy, but the spring starts to stretch. At the bottom, when the spring has absorbed all of the person's kinetic energy, all of the change in gravitational energy has transformed into spring energy. That's why here, you equate the energies.

In your first example, it was like an external influence had come in and pulled the person up to a position where the net force would be zero, and the person stayed there. That place just happens to be right in between the highest and lowest points. That's where the factor of two arises. In the second question, the person moves.

[SocialRhubarb]

This is slightly different. At the bottom of his bungee jump, Sam is momentarily at rest, meaning that his kinetic energy is 0. Hence, it is valid for us to equate the change in gravitational potential to the change in elastic potential energy.

The difference between this question and the previous question is the starting point of the system.

In theory, assuming ideal springs and no retarding forces, if you had a ball on a spring and you set it going, it would never come to rest for more than an instant. From this conclusion, we conclude that the object in the first question must have started in its resting position, and hence no change in gravitational potential energy has occurred.

However, looking at the bungee jumper, he started far above the equilibrium position, which is the point at which the spring constant is equal to the weight force. That means that by the time he reaches this equilibrium position, he has quite a lot of kinetic energy. That is to say, the question states that Sam momentarily stops at 18m, but the point at which he momentarily stops is not the point at which his weight force is equal to the force from the spring. At the point which weight force is equal to the spring's force, his acceleration is 0, but his velocity is actually at a maximum, and so the point at 18m where Sam is momentarily stationary is not when mg=kx.

[iGoRaw]

Ok. This is the original question.
It says in the stem that when the mass is attached the spring and mass are stationary at a length of 70 cm.
So stationary starts and ends. At 70cm there should be no Kinetic energy right? So why can't we equate SPE and GPE?

[lzxnl]

Because if you just attach a spring, it WON'T be stationary at 70 cm. Although the net force is zero, as SocialRhubarb points out, the object has kinetic energy, so it keeps going down. Some external force is necessary to reduce the kinetic energy to zero at the point of equilibrium.
This is not an energy question; it's a forces question.

[iGoRaw]

Ok..
So with springs, when vertically hanged always use F=mg, F=kx to solve for K?

[lzxnl]

Yes. Use forces as the net force is zero.