# Conservation of energy problem

1. Aug 20, 2014

### Curieuse

1. The problem statement, all variables and given/known data

The figure shows an 8.00 kg stone at rest on a spring. The spring is compressed by 10.0 cm by the stone.
What is the spring constant?

2. Relevant equations

Elastic potential energy= $\frac{k{x^{2}}}{2}$
Gravitational potential energy= mgh
Spring force F= -kx

3. The attempt at a solution

Okay, so I solved this using a free body diagram and newton's laws
kx = mg -------------------------------------(1)
So, k=(mg)/x= 784 N/m
And this is the answer, alright!

But then i tried using Energy considerations,

And i did this:

Increase in the elastic potential energy of the spring-stone system = decrease in the gravitational potential energy of the stone-earth system

So, $\frac{k{x^{2}}}{2}$ = mgx
which simplifies to give $\frac{kx}{2}$ = mg -------------------------- (2)
and this is obviously wrong!
Where did i go wrong? Is there some other energy i should've considered?
I ruled out Kinetic energy because the system is stationary.
The half term in the l.h.s of (2) is what is causing the problem, evidently! Please help!

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2. Aug 20, 2014

### Staff: Mentor

If you just drop the mass on the unstretched spring you will have kinetic energy as well as spring potential energy to consider. Here, the mass is gently placed on the spring, not dropped, so the missing energy went into whatever helped lower the mass.

The correct way to solve this is using a force analysis, as in your first method.

3. Aug 20, 2014

### Curieuse

and that is?

4. Aug 20, 2014

### Curieuse

how about i consider a force F such that for x=0 to x= 10 cm, this force F makes sure that there isn't any acceleration in the mass, it adjusts itself accordingly as F= mg-kx ?
Then i do get the answer! But is this alright?

5. Aug 20, 2014

### Staff: Mentor

Sure, that's fine. And when the spring is at rest at the equilibrium point, that applied force is 0 and you're back to just mg = kx. So why not start there?

I would say that this is not a conservation of energy problem, if that's what you're still thinking. A typical conservation of energy type problem would be to place the mass on the unstretched spring and just drop it. Given k, what will be the maximum compression of the spring?

6. Aug 20, 2014

### Staff: Mentor

Depends on how the mass was lowered onto the spring. Could just be your hand!

7. Aug 20, 2014

### Curieuse

Yes, true that. HRW listed it in that chapter, so i wondered if those rules could be applied here too! but why is it tricky? isnt there an easier way to do it using these laws? without considering any more F forces?

8. Aug 20, 2014

### Orodruin

Staff Emeritus
Well, you could approach the problem through energy conservation and the knowledge that the spring-stone system is going to be a harmonic oscillator by arguing along the following lines:

You know the potential energy at the release point (0 for both spring and gravity). At the turning point of the other side of the equilibrium, you will have no kinetic energy and so again the potential must be zero -- this is in fact what you solved for above - obtaining $x = 2mg/k$. Since the oscillator is harmonic, the equilibrium point must be exactly in between the two turning points, at $x = mg/k$, which is the same result as the one you obtained in your force analysis.

9. Aug 20, 2014

### Staff: Mentor

What chapter?

Easier than what? You are told that it's sitting there at equilibrium. And you solved it easily by considering forces.

Energy conservation isn't particularly useful here. But it's important to understand why it doesn't apply. Energy conservation usually connects one time/position to another. Here you just have it sitting there.

10. Aug 20, 2014

### Curieuse

Conservation of energy

11. Aug 20, 2014

### Staff: Mentor

So you are saying that this problem (from your initial post) appeared in the conservation of energy chapter? Were there other parts to it?

12. Aug 20, 2014

### Curieuse

yes! chapter 8! but in the instructor's manual, they used the F= kx=mg, so yeah!

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13. Aug 20, 2014

### Orodruin

Staff Emeritus
Well, the problem has several other parts that should be approached with energy considerations. In general, the fact that you are in a different chapter does not mean you should forget what you learned earlier - in fact it is very common that new things build upon knowledge already acquired.

14. Aug 20, 2014

### Curieuse

no no i didn't forget.. i just wanted to try another approach and see if it would work as well!

15. Aug 20, 2014

### Orodruin

Staff Emeritus
Well, technically you could do it the way I described before. Or you could write down the potential and set the derivative to zero. In the end, the approaches are just variants of the same thing.

16. Aug 20, 2014

### Curieuse

Yep! I overlooked your first post! Thanks a lot!!

17. Aug 20, 2014

### Staff: Mentor

Sure, it's just the first part of the problem. You'll then use that result and apply energy methods to solve the other parts.

18. Aug 20, 2014

### Curieuse

Yeah exactly that's why i didn't mention the rest!