# Spring falling

1. May 16, 2009

### Myvalq

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

A mass (weight 100g) is hanged on a spring inside a cubic box (side a=1m). Stiffness of the spring is 10N*m-1. Equilibrium state of the spring is right in a centre of the box.
Now, we let the box fall to the ground from height h. Describe a behaviour of the mass inside the box. We can assume that the box is very heavy and the spring is an ideal one.

2. Relevant equations

3. The attempt at a solution

2. May 16, 2009

### Dr.D

Maybe you should make a start on this problem first. Try drawing a FBD for the mass shortly after the box has hit the ground and see what you can do with that. Choose some coordinates, etc.

3. May 17, 2009

### Shooting Star

I think that a description of the behaviour of the system before it hits the ground has been asked for. Since everything is accelerating downward at g, there is effectively no gravity in the frame of the box, and the spring is weightless -- only tension is acting on the spring. Now can you describe what happens to the mass?

4. May 18, 2009

### Myvalq

Firstly, while falling, there appears a weightlessness=>no gravity downward. But then there is no equillibrium state because nothing acts against the elastic force of the spring. And therefore the mass goes up until the spring is completely scrolled.
Is that right?
And what happened next(after it hits the ground)? (Can someone help me pls, I am desperate)

5. May 18, 2009

### Shooting Star

There may not be any effective gravity in the box, but the sping was extended and it will try to come back to its unstretched position -- just like a stretched spring with a mass attached resting on a horizontal table. (There is nothing to dissipate the energy of the mass.) What happens to the spring-mass system now?

After it hits the ground, the box comes to rest immediately whereas the the mass will tend to move downward. What happens to the spring-mass system now?

6. May 18, 2009

### Myvalq

So while falling,, the spring comes back to its unstretched position (=equillibrium state?) and there stays until it hits the ground?

And then some inertial force begins to act? (what to do with the height?)

7. May 18, 2009

### Shooting Star

The mass oscillates about the mean position. The mean position is given by the unstretched length of the spring and the extreme position is given by the stretched length.

After the box hits the ground, the mass oscillates about a new mean position. This mean position of the mass is given by the length of the spring after the spring is stretched by the weight of the mass. Now try to figure out the maximum amplitude.

8. May 19, 2009

### Myvalq

Why does it oscillate during the fall? I don't see any force acting downwards?

When on the ground: The initial energy of the mass is E=mgh (where h is the height) and when spring goes down it changes to the energy of elasticity: E=(1/2)ky2.
And when the spring is in the downmost position the E of elesticity is biggest: mgh=(1/2)ky2 => y=$$\sqrt{2mgh/k}$$ and and that is the maximum amplitude.(?)

Last edited: May 19, 2009
9. May 20, 2009

### Shooting Star

Read post no. 5 again. A spring tends to stretch when compressed.
What is h here? It's better if you measure distance from the equilibrium point of the mass. The mean position of the mass is different when gravity is zero from the mean position when there is gravity.

(I still feel that the question only asks for how the mass behaves when in free fall.)

10. May 24, 2009

### stewdonym

Can you draw the force diagram that shows this (the different forces acting on the mass as the box is falling)?

Last edited: May 24, 2009