Compression of a spring. Hooke's Law vs. Conservation of Energy

In summary, the conversation discusses the compression of a spring with a stiffness of 250 N/m when a 5 kg block is placed on top of it. Two different methods are used to calculate the compression, one using Hooke's Law and the other using conservation of energy. The correct answer is determined to be the first one, as the second method does not take into account the kinetic energy developed by the mass. The conversation ends with a clarification on the equilibrium point and the resulting oscillation of the mass.
  • #1
thindelgado
3
0

Homework Statement



Lets say I have a spring with a stiffness of k = 250 N/m originally unstretched. I then gently place a 5 kg block on top of the spring. How much does the spring compress?


Homework Equations



W = mg
F = -ks
mgΔh = 0.5k(Δh)2

The Attempt at a Solution



Using Hooke's Law I get the following result

W = 5*9.81 = 49.05 N

W = F = -kx

49.05 = 250x meaning x = 0.1962 m

Using conservation of energy I get this though:

mgx = 0.5kx2

0 = 125x2 - 49.05x

x = 0.3924 m

So which one is the correct answer? Why is one double the other?
 
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  • #2
The first answer is correct. If you place it gently on the spring, then you need to apply force (and do work) to keep it from accelerating while the spring is compressing. If you don't do this, then the mass will have kinetic energy as it passes through the equilibrium point, and it won't stop until it has traveled twice as far. But this will not be an equilibrium point, and the mass will go back up in the other direction. If you apply the force to keep it from accelerating, then it will only go down to the equilibrium point x = mg/k. The problem with your energy balance was that you omitted the kinetic energy developed by the mass.

This same question has appeared on Physics Forums many times in the past.

Chet
 
  • #3
Chestermiller said:
The problem with your energy balance was that you omitted the kinetic energy developed by the mass.

All right, here's my confusion though. If the block is originally at rest then it has no kinetic energy only gravitational potential. Once it starts descending (compressing) it has some kinetic but when it reaches the lowest it can get, for that moment all of its energy is in the form of spring potential. So why wouldn't the initial gravitational potential energy be equal to the spring potential energy?
 
  • #4
The "lowest you can get" is not the same as the position reached when you "gently place it".
It is actually twice as much.
You are calculating two different things so you get two different answers.
When you place it gently the energy of the weight+spring system is not conserved.You have some external force doing work on the system, as was already explained by Chet.
 
  • #5
Ok, I think I'm getting it now. This would be my last counterargument I guess.

So if the block was released from rest 1 m above the spring then with conservation of energy:

mg(h+x) = .5kx^2
49.05 + 49.05x = 125x^2
x = .8526 m

I'm assuming this is correct since pretty much the same scenario is in an example in my book.

So again, if i was to release it from rest when h = 0, then I would go back to mgx = 0.5kx^2. How has anything changed?

Thank you guys for the help!
 
  • #6
thindelgado said:
All right, here's my confusion though. If the block is originally at rest then it has no kinetic energy only gravitational potential. Once it starts descending (compressing) it has some kinetic but when it reaches the lowest it can get, for that moment all of its energy is in the form of spring potential. So why wouldn't the initial gravitational potential energy be equal to the spring potential energy?

It will be equal to the spring potential energy. But this will not be the equilibrium point. When you get to this point, the spring will have been compressed to twice the distance to the equilibrium point, and the upward force of the spring will be 2mg instead of mg. So the mass will accelerate upward again. The mass will forever oscillate up and down in simple harmonic motion from zero displacement to twice the equilibrium displacement. Whenever it passes through the equilibrium point, its velocity will not be zero; it will pass through a maximum.

Chet
 

1. What is Hooke's Law and how does it relate to the compression of a spring?

Hooke's Law is a principle in physics that states that the force required to compress or stretch a spring is directly proportional to the distance the spring is compressed or stretched. This means that the more you compress a spring, the more force it will exert to return to its original length. This law applies to the compression of a spring because as the spring is compressed, the molecules within the spring are pushed closer together, causing an increase in force.

2. How does the conservation of energy apply to the compression of a spring?

The conservation of energy states that energy cannot be created or destroyed, only transferred or transformed. This applies to the compression of a spring because when a spring is compressed, the potential energy stored within the spring is transformed into kinetic energy, which causes the spring to bounce back to its original shape.

3. Can you explain the relationship between Hooke's Law and the conservation of energy?

The relationship between Hooke's Law and the conservation of energy is that they both describe different aspects of the same physical phenomenon - the compression of a spring. Hooke's Law describes the relationship between force and displacement, while the conservation of energy describes the transformation of potential energy into kinetic energy.

4. How do you calculate the potential energy of a compressed spring?

The potential energy of a compressed spring can be calculated using the equation PE = 1/2kx^2, where PE is the potential energy, k is the spring constant, and x is the distance the spring is compressed. This equation is derived from Hooke's Law and is based on the principle that the potential energy stored in a spring is directly proportional to the force required to compress it.

5. Can Hooke's Law be applied to all types of springs?

Yes, Hooke's Law can be applied to all types of springs, including coil springs, leaf springs, and torsion springs. However, the spring constant (k) may vary depending on the type of spring and its material properties. It is important to note that Hooke's Law is only applicable within the elastic limit of a spring, meaning that once the limit is exceeded, the spring will no longer obey the law and may become permanently deformed.

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