# Spring Conservation with Gravity

• Ronnin
In summary, the spring with a negligible mass and a 400N/m force constant pushes a .2kg pan with an inelastic collision quickly into a harmonic oscillation.
Ronnin
A spring with negligible mass and force constant of =400N/m is vert and a .2kg pan is suspended from its lower end. A butcher drops a 2.2kg steak onto the pan from a height of .4m. It makes a totally inelastic collision w/ the pan and sets the system into SHM.

A) What is the speed of the pan and steak immediatly after collision

Kf + Uf=Ki + Ui
v=(2gh)^1/2
v=2.8m/s

Vf=(m/M)Vm
(2.2/2.4)2.8=2.57m/s

This part I had no problems with, now

B) The amp of the subsequent motion

I can figure the KE of the system, but I know I must account for gravity in the PE as well as the spring's force. I know the compression of the spring due to gravity is mg/k which is .059m.

Spring PE=.5kx^2=.696J
KE=.5(2.4)2.57^2=7.926J

But a solution I saw actually subtracted the spring PE from the KE and got 7.34J total energy for the system. I thought total energy was PE + KE

There is in fact GPE involved in this problem. You can use it and the initial kinetic energy to determine the lowest point the steak will reach. If you then find the equilibrium point, you can deduce the amplitude of the harmonic motion.

There is another way to look at it. The gravitational force is constant in the problem. Its effect is to move the equilibrium point of the steak/spring system and to establish that point as the zero of elastic potential energy. Remember that the zero point of GPE is arbitrary. Immediately after impact, the steak is above equilibrium with the kinetic energy you calculated. When the spring reaches the equilibrium point, the steak will be moving more slowly than it would be otherwise because the spring is stretched. The kinetic energy at that point is what will be defined as the total energy of the system, so that the elastic energy (really the combined spring/gravity potential energy) is zero at that point. The steak loses some amount of kinetic energy that is transferred to the spring, but it also gains kinetic energy because of the additional loss of GPE. I doubt that the calculation you saw was correct if it left out the GPE, but I can see the motivation for subtracting spring energy.

I suggest you do the problem the way you are thinking because although there are alternatives I think you will find the total energy of the system relative to the equilibrium point is actually greater than the kinetic energy after impact. You know that it has to be if the subsequent motion is harmonic because the greatest kinetic energy of a harmonic oscillator is at the equilibrium point. At impact the steak is still above equilibrium point. It is going to go faster before it gets there.

Be very careful about the distances. Don't foget the spring is already stretched by the pan before the steak arrives.

Last edited:
So it is not until my new equilibrium point that the KE from the falling weight starts applying to the PE of the spring? So I should consider all things zero at the new equilibrium point with the weight applied?

Ronnin said:
So it is not until my new equilibrium point that the KE from the falling weight starts applying to the PE of the spring? So I should consider all things zero at the new equilibrium point with the weight applied?
That does not sound like what I am saying As soon as the steak and the pan merge into one they have the kinetic energy you calculated. The spring was already slightly stretched by the pan before the steak hit, so it already had some stored energy. Now the steak/pan continues to move downward, losing GPE, increasing the elastic energy of the spring, and changing KE with the sum of these three energies remaining constant. At the equilibrium point we can choose to call the total potential energy (GPE plus spring energy) zero. At this point the steak/pan have maximum speed.

If you figure out where the equilibrium point is, you could choose your reference point for GPE so that at equilibrium the GPE is the negative of the spring energy so the total potential energy is zero. I am not suggesting you actually do that, but you could. The important thing is that you figure out how far from equilibrium the steak/pan comes to rest, using whatever reference point for GPE is convenient for you.

Since gravity is a constant force, and the spring force is proportional to displacement, it turns out that the net force from gravity and the spring combined acting on the steak/pan is proportional to the displacement from equilibrium. This is exactly the condition needed for simple harmonic motion. You no longer need to think in terms of GPE and spring energy as separate quantities. Their combined effect can be written as an "elastic-like energy". If x is the displacement from equilibrium this potential energy is the familiar form ½kx². At maximum x, KE is zero. At x = 0, KE is max. The total energy of the system is conserved.

Last edited:

## 1. What is "Spring Conservation with Gravity"?

"Spring Conservation with Gravity" refers to the conservation of energy and momentum in a system involving a spring and gravity. This concept is commonly used in physics and engineering to analyze the motion of objects connected to a spring in the presence of gravitational forces.

## 2. How is energy conserved in a spring with gravity system?

In a spring with gravity system, energy is conserved through the conversion of potential energy (stored energy) into kinetic energy (energy of motion). As the spring stretches or compresses, potential energy is stored in the spring. When the spring is released, this potential energy is converted into kinetic energy, causing the object connected to the spring to move.

## 3. What is momentum conservation in a spring with gravity system?

Momentum conservation in a spring with gravity system refers to the principle that the total momentum of the system (object connected to the spring and the spring itself) remains constant as long as there are no external forces acting on the system. This means that the momentum gained by the object will be equal to the momentum lost by the spring, and vice versa.

## 4. How does gravity affect the motion of a spring?

Gravity affects the motion of a spring by exerting a downward force on the object connected to the spring. This force, known as weight, can cause the spring to stretch or compress depending on the direction of the force and the stiffness of the spring. Gravity also contributes to the potential energy in the system, which is converted into kinetic energy as the spring moves.

## 5. What real-life applications use "Spring Conservation with Gravity"?

There are many real-life applications that use "Spring Conservation with Gravity," such as bungee jumping, trampolines, and car suspension systems. These systems all involve the use of a spring and gravity to store and release energy, resulting in controlled and efficient motion. Engineers also use this concept when designing structures or machinery that need to withstand gravitational forces.

### Similar threads

• Introductory Physics Homework Help
Replies
20
Views
2K
• Introductory Physics Homework Help
Replies
24
Views
983
• Introductory Physics Homework Help
Replies
12
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
783
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
354
• Introductory Physics Homework Help
Replies
9
Views
8K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
15
Views
3K