Mass in a Falling Box: Investigating its Behaviour

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The discussion revolves around the behavior of a mass attached to a spring inside a falling box. While the box is in free fall, the mass experiences a state of weightlessness, leading to no gravitational force acting on it, which allows the spring to extend as it tries to return to its unstressed position. Upon impact with the ground, the box stops abruptly, but the mass continues to move downward due to inertia, causing oscillation around a new mean position determined by the spring's stretch. The energy transformation from gravitational potential energy to elastic potential energy is also discussed, highlighting the maximum amplitude of oscillation. The analysis emphasizes the importance of understanding forces and energy changes in such dynamic systems.
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Homework Statement



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.

Homework Equations





The Attempt at a Solution

 
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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.
 
Myvalq said:

Homework Statement



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

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?
 
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)
 
Myvalq said:
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)

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?
 
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?)
 
Myvalq said:
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?)

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.
 
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.(?)
 
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Myvalq said:
Why does it oscillate during the fall? I don't see any force acting downwards?
Read post no. 5 again. A spring tends to stretch when compressed.
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.(?)
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
Shooting Star said:
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?

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