Conservation of energy - spring

In summary, the problem involves finding the amount of compression on a spring when a 5kg weight is dropped from a height of 0.4m, with a spring elasticity factor of 1700N/m. The attempt at a solution involved setting the potential energies when the weight is dropped and when it reaches the bottom equal to each other, resulting in a quadratic equation. However, the correct answer may be the positive solution, as the negative solution may not make sense in this context. It was also noted that the gravitational potential energy term may have been incorrect, and should include the acceleration due to gravity.
  • #1
lukatwo
24
0

Homework Statement


hh86igL.png

Hello, I'm having a problem with this. So we drop the weight on this spring, and the question is: how much does the spring compress? The mass is 5kg, the height 0,4m, and the spring elasticity factor k=1700N/m.

Homework Equations





The Attempt at a Solution


I've tried to equalize the potential energies of when we drop the weight, and when it's all the way down. It looks like this: m(h+ΔL)=(1/2)*k*(ΔL)^2. I get a quadratic equation, and solve it. I get two answers one with +, and one -. I've tried to enter the positive one, and it's not correct. Should I enter the negative one, or is my whole attempt faulty?
 
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  • #2
lukatwo said:

Homework Statement


hh86igL.png

Hello, I'm having a problem with this. So we drop the weight on this spring, and the question is: how much does the spring compress? The mass is 5kg, the height 0,4m, and the spring elasticity factor k=1700N/m.

Homework Equations





The Attempt at a Solution


I've tried to equalize the potential energies of when we drop the weight, and when it's all the way down. It looks like this: m(h+ΔL)=(1/2)*k*(ΔL)^2. I get a quadratic equation, and solve it. I get two answers one with +, and one -. I've tried to enter the positive one, and it's not correct. Should I enter the negative one, or is my whole attempt faulty?
Looks like you left out g in your gravitational PE term, otherwise, looks good. Think positive.
 
  • #3
Should it be g*m*(h+∆L)=(1/2)*k*(∆L)^2, because if it's on both sides it's the same
 
  • #4
lukatwo said:
Should it be g*m*(h+∆L)=(1/2)*k*(∆L)^2, because if it's on both sides it's the same
Sure!
 
  • #5


I would like to clarify that conservation of energy is a fundamental principle in physics which states that energy cannot be created or destroyed, only transferred or converted from one form to another. In the scenario described, the energy of the weight is converted into potential energy as it is lifted to a height of 0.4m, and then into elastic potential energy as it compresses the spring.

To solve for the amount of compression of the spring, we can use the equation for elastic potential energy: Ee = (1/2)*k*(ΔL)^2, where Ee is the elastic potential energy, k is the spring constant, and ΔL is the change in length of the spring.

Substituting the given values, we get: Ee = (1/2)*1700*(ΔL)^2 = (1/2)*5*9.8*0.4 = 9.8J

Solving for ΔL, we get ΔL = √(9.8/850) = 0.041m.

Therefore, the spring will compress by 0.041m.

Your attempt at using the potential energy equation is correct, but it seems that you may have made a mistake in your calculations or in setting up the equation. I would suggest double-checking your work and trying again. It is possible that the negative answer you obtained is the correct one, but it is always important to check your work and make sure it aligns with the principles of conservation of energy.
 

1. What is the concept of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, but can only be transformed from one form to another. This means that the total amount of energy in a closed system remains constant.

2. How does a spring demonstrate conservation of energy?

When a spring is compressed or stretched, it stores potential energy. This potential energy is then converted into kinetic energy as the spring returns to its original shape. The total energy remains the same, but it is transformed from potential to kinetic and vice versa.

3. Can energy be lost in a spring?

No, energy cannot be lost in a spring. As mentioned before, the total energy in a closed system remains constant. When a spring is compressed or stretched, the potential energy may decrease but it is converted into kinetic energy. Therefore, the total energy remains the same.

4. How does the mass of an object affect the conservation of energy in a spring?

The mass of an object does not affect the law of conservation of energy in a spring. The amount of energy stored in a spring depends on its displacement and spring constant, not on the mass of the object attached to it.

5. Can the conservation of energy be applied to real-life situations involving springs?

Yes, the law of conservation of energy can be applied to real-life situations involving springs. For example, in a car's suspension system, the energy from bumps on the road is absorbed by the springs and then released as the car moves over the bump. This demonstrates the conversion of potential to kinetic energy in a spring system.

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