Energy (spring) Question

In summary, the question asks for the maximum elongation of a spring with a spring constant of 200N/m, when a 2-kg block is given a speed of 5 m/s at the spring's equilibrium point. This can be solved using the conservation of energy principle.
  • #1
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I have a question involving the spring costant:

A 2-kg block is attached to a horizontal ideal spring with a spring constant of 200N/m. When the spring has its equilibrium length the block is given a speed of 5 m/s. What is the maximum elongation of the spring?

To my understanding when the spring is at its equilibrium length, isn't there no force acting on the spring. So how would you work around it to solve for length. I would appreciate help in how to go about solving the problem. Thank you.
 
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  • #2
blackout85 said:
I have a question involving the spring costant:

A 2-kg block is attached to a horizontal ideal spring with a spring constant of 200N/m. When the spring has its equilibrium length the block is given a speed of 5 m/s. What is the maximum elongation of the spring?

To my understanding when the spring is at its equilibrium length, isn't there no force acting on the spring. So how would you work around it to solve for length. I would appreciate help in how to go about solving the problem. Thank you.
That is a bit confusing question. I'm guessing it means that the spring is stretched initially, then released, and the block has a speed of 5m/s as it passes the equilibrium point of the spring. Use the conservation of energy principle
initial KE of block plus initial PE of spring = final KE of block plus final PE of spring. A couple of those terms are 0.
 
  • #3


Dear questioner,

Thank you for your question regarding the maximum elongation of a spring. I am happy to assist you in solving this problem.

Firstly, it is important to understand that a spring's equilibrium length is the point at which it experiences no net force. This means that the forces acting on the spring are balanced and it is not moving. However, when an external force is applied to the spring, it will stretch or compress from its equilibrium length.

In this scenario, the 2-kg block is given a speed of 5 m/s, which means it has kinetic energy. As the block is attached to the spring, this kinetic energy will be transferred to the spring, causing it to stretch from its equilibrium length. The maximum elongation of the spring will occur when all of the block's kinetic energy is transferred to the spring and the block comes to a stop.

To solve for the maximum elongation, we can use the conservation of energy principle. This states that the total energy in a closed system remains constant. In this case, the system includes the block, the spring, and the Earth's gravitational field.

We know that the initial energy of the system is the kinetic energy of the block, which is given by the equation KE = 1/2 * m * v^2, where m is the mass of the block and v is its velocity. In this case, the initial kinetic energy is 1/2 * 2 kg * (5 m/s)^2 = 25 J.

As the block comes to a stop, all of this energy is transferred to the spring in the form of potential energy. The potential energy of a spring is given by the equation PE = 1/2 * k * x^2, where k is the spring constant and x is the displacement from the equilibrium length. We can rearrange this equation to solve for x, the maximum elongation of the spring, which gives us x = √(2PE/k).

Substituting in the values, we get x = √(2*25 J/200 N/m) = 0.5 m. This means that the maximum elongation of the spring is 0.5 m.

I hope this explanation helps you understand how to solve for the maximum elongation of a spring. If you have any further questions, please don't hesitate to ask. Thank you for your interest in science.

Sincerely,
[Your
 

1. What is potential energy in a spring?

Potential energy in a spring is the energy stored in the spring when it is stretched or compressed. This energy is stored in the form of elastic potential energy and is dependent on the spring's stiffness and how far it is stretched or compressed from its equilibrium position.

2. How is the potential energy of a spring calculated?

The potential energy of a spring can be calculated using the equation PE = 1/2kx^2, where k is the spring constant and x is the displacement from the equilibrium position. This equation is derived from Hooke's law, which states that the force exerted by a spring is directly proportional to its displacement.

3. What is the relationship between the potential energy and the displacement of a spring?

The potential energy of a spring is directly proportional to the square of its displacement from its equilibrium position. This means that as the displacement increases, so does the potential energy stored in the spring. When the spring is at its equilibrium position, it has no potential energy.

4. How does the stiffness of a spring affect its potential energy?

The stiffness of a spring, represented by the spring constant k, directly affects the potential energy stored in the spring. A stiffer spring (higher k value) will have a greater potential energy at the same displacement compared to a less stiff spring (lower k value).

5. How is the potential energy of a spring converted into other forms of energy?

The potential energy stored in a spring can be converted into other forms of energy, such as kinetic energy, thermal energy, or sound energy. This conversion occurs when the spring is released and the stored energy is released, causing the spring to vibrate or move. The amount of energy converted depends on the stiffness of the spring and the distance it is stretched or compressed.

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