Spring constant discrepency (force vs work)

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Homework Help Overview

The discussion revolves around the discrepancy in the spring constant derived from two different approaches: one based on the work done to stretch a spring and the other based on the force exerted by the spring. The subject area is mechanics, specifically focusing on spring dynamics and gravitational forces.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the work done by gravity and the work required to stretch a spring, questioning why different constants arise from these calculations. Some participants discuss the implications of kinetic energy and the nature of forces involved in stretching the spring.

Discussion Status

The discussion is active, with participants raising questions about the nature of work done on the spring and the energy transformations involved. Some guidance is provided regarding the relationship between force and work in the context of spring mechanics, but no consensus has been reached.

Contextual Notes

Participants note that the work done by gravity does not equal the work to stretch the spring, highlighting a potential misunderstanding of energy distribution in the system. The discussion also touches on the assumptions made regarding the behavior of the mass and spring system.

krausr79
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y = distance stretched vertically
k = spring constant
M = mass of weight on spring

My Physics book says the work to stretch a spring is .5ky2. If I set that equal to the work done by gravity on the weight (9.8xM) and solve for the constant I get k = (9.8M)/(.5y)

My Physics book also says that the force from a spring is ky. If I set that equal to the force of gravity on the weight (9.8M) and solve for the constant I get k = (9.8M)/(y)

Why do I get a different constant??
 
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krausr79 said:
My Physics book says the work to stretch a spring is .5ky2. If I set that equal to the work done by gravity on the weight (9.8xM) and solve for the constant I get k = (9.8M)/(.5y)
But the work done by gravity does not equal the work to stretch the spring--it's twice as much as needed to just stretch the spring.

My Physics book also says that the force from a spring is ky. If I set that equal to the force of gravity on the weight (9.8M) and solve for the constant I get k = (9.8M)/(y)
That's true. It's a restatement of Hooke's law.
 
That would explain it. But where does the extra work go? Is that the right question? Why would only half go to the spring?
 
krausr79 said:
That would explain it. But where does the extra work go? Is that the right question? Why would only half go to the spring?
The extra work goes into added kinetic energy of the mass.

Here's what's going on. To just barely stretch the spring, you only need to exert a force equal to the spring force. But that spring force varies from zero (at first) to the full force (mg) at the equilibrium point. So the work needed--Average Force x distance--is only ½(mg)y = ½ky².

If you gently hang the weight on the spring and bring it down to full extension realize that you are countering the weight of the mass with the upward force of your hand. But if you hang the weight on the spring and just let it drop, then the work done by gravity (when the weight reaches the point y) will be mgy = ky^2, but that extra energy goes into speeding up the weight. It won't just stop, it will keep going until the spring stretches enough to stop the motion. Ignoring friction and other energy leaks, the weight will keep oscillating up and down in simple harmonic motion.
 
Thank you.
 

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