How Does Kinetic Energy Affect Spring Elongation?

[blackout85]

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.

 

[PhanthomJay]

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.

 

[kyle8921]

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,
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