How Does Kinetic Energy Affect Spring Elongation?

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SUMMARY

The discussion centers on calculating the maximum elongation of a spring when a 2-kg block is attached to it and given an initial speed of 5 m/s. The spring has a spring constant of 200 N/m. To solve for the maximum elongation, the conservation of energy principle is applied, equating the initial kinetic energy of the block and the potential energy stored in the spring at maximum elongation. The key takeaway is that the spring's elongation can be determined by balancing kinetic and potential energy at the equilibrium point.

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blackout85
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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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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.
 

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