# Stupidity at its peak find the mass

1. Jul 27, 2011

### flyingpig

1. The problem statement, all variables and given/known data
A spring is hung from a ceiling, and an object attached to its lower end stretches the spring by a distance of 5.00 cm from its unstretched position when the system is in equilibrium. If the spring constant is 43.2 N/m, determine the mass of the object.

3. The attempt at a solution

$$\sum W = \Delta K$$

$$mg\Delta y - \frac{k}{2}(y^2) = 0$$

$$ky = 2mg$$

I will stop here because this is wrong.

Apparently using Newton's forces

kx = mg

There is no 2...

2. Jul 27, 2011

### ehild

The object is equilibrium, so the net force on it is ??

ehild

3. Jul 27, 2011

### flyingpig

0!

Why am I wrong though? That's what I want to know.

4. Jul 27, 2011

### ehild

And what forces act on the object?

ehild

5. Jul 27, 2011

### flyingpig

Spring and the mysterious force that's pulling it down that is also equal to the (positive) work done by gravity

6. Jul 27, 2011

### ehild

I start to understand you. It does not matter, how the object is brought to equilibrium. Maybe, you hold the object and let it move very slowly till it reaches equilibrium. Then the net work, yours and that of the spring and that of gravity is zero and you do not reach anywhere. You need to calculate with the forces. So what forces act on the object, hanging at the end of the spring when it does not move any more?

ehild

Last edited: Jul 27, 2011
7. Jul 27, 2011

### flyingpig

Well it went down, oh wait it i just gravity. Not so mysterious now...

8. Jul 27, 2011

### ehild

You see...:rofl:

ehild

9. Jul 27, 2011

### flyingpig

Yeah and I added them, but it didn't give the same answer as Newton's forces?

10. Jul 27, 2011

### ehild

What are Newton's forces, and what did you get for the result?

ehild

11. Jul 27, 2011

### flyingpig

kx = mg

with newton's forces

12. Jul 27, 2011

### ehild

Well, is not this the same as saying that kx force acts up and mg down, so the resultant force is kx-mg and it is zero as the object is in rest? So what is the mass of the object?

ehild

13. Jul 27, 2011

### flyingpig

kx = mg

m = kx/g

But I still don't understand why I am wrong I thought I did it correctly.

14. Jul 27, 2011

### ehild

In what are you wrong?

ehild

15. Jul 27, 2011

### flyingpig

With energy I got kx = 2mg which is wrong.

16. Jul 27, 2011

### ehild

It is wrong as the spring never reaches equilibrium if only the spring force and mg act on it. When the object connected to the unstretched spring and released it will oscillate around the equilibrium point, with the distance x=mg/k below the initial position. Its KE is maximum here, insted of zero as you assumed. Try to write Newton's second law for the acceleration and solve the differential equation for x.

ehild

17. Aug 3, 2011

### flyingpig

Is the question suggesting we take a particular instant then when you say it keeps oscillating back and forth? Why can't we take the work done at a particlular instant here?

18. Aug 3, 2011

### ehild

The question states equilibrium.

Being in equilibrium means rest; zero velocity, zero acceleration, zero net force. An oscillating object is not in equilibrium.

When the object is released after attached to the end of the unstretched spring, the spring will stretch by ΔL=2mg/k when the velocity of the object is zero again. At that instant, an upward force F=mg acts on the object. The acceleration is upward, the object moves upward. At ΔL=mg/k, the net force is zero, but the object has velocity, so it will move upward further, till ΔL=0, where the force is downward: The object will oscillate between ΔL=0 and ΔL=2mg/k.

ehild