Work energy question. Rock down a hill.

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Homework Help Overview

The discussion revolves around a physics problem involving a fish attached to a vertical spring and its behavior when lowered to equilibrium and then dropped. The subject area includes concepts from mechanics, specifically the work-energy theorem and spring dynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between gravitational potential energy and spring potential energy, questioning the correct application of the work-energy theorem. There are attempts to derive the maximum stretch of the spring when the fish is dropped, with varying interpretations of the variables involved.

Discussion Status

Multiple interpretations of the problem are being explored, with participants offering different equations and reasoning. Some guidance has been provided regarding the correct form of the energy equations, but there is no explicit consensus on the final outcome or correctness of the approaches.

Contextual Notes

Participants are grappling with the implications of the spring's stretch and the relationship between the distance the fish falls and the spring's equilibrium position. There is a hint of confusion regarding the definitions of variables and their physical meanings in the context of the problem.

Exuro89
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Work energy question. Fish on a spring.

Question has been changed as I figured it out. New one is on fish and springs

Homework Statement


If a fish is attached to a vertical spring and slowly lowered to its equilibrium position, it is found to stretch the spring by an amount d.
If the same fish is attached to the end of the unstretched spring and then allowed to fall from rest, through what maximum distance does it stretch the spring? (Hint: Calculate the force constant of the spring in terms of the distance d and the mass m of the fish.)

Homework Equations


Work Energy Theorem
F=-kx



The Attempt at a Solution


I understood the hint, in which the force constant is k = (mg)/d

Now I need to come up with a work energy equation. Energies would be potential spring and gravitational. Initial would be only gravitational as the spring hasn't been stretched. The final energy should be all spring. So the equation would be
0 = -mgh + 1/2kd^2 or mgh = 1/2kd^2

h is the maximum distance the spring will go. I need to replace k with (mg)/d so I only have those few variables, so

mgh = 1/2(mg/d)d^2

mgh = 1/2mgd

h = 1/2d

Well I'm pretty sure this is wrong. The total distance should be greater than the stretch no? What is it that I'm doing incorrectly?
 
Last edited:
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I think I got it. Can someone check this?

So my potential spring energy is incorrect I believe. In this problem the mass would be stretching the spring out to x length and not d length so the equation should look like this.

mgx = 1/2kx^2

Since k = mg/d I can sub that in.

mgx = 1/2(mg/d)x^2

mg cancels out

x = 1/2d*x^2

move 1/2d over

2dx = x^2

divide out x

2d = x

This looks plausible. Since d would be the new equilibrium another d in length would be stretched if the mass was dropped with the spring connected. Is this correct?
 
Does animal right activists know about all this?
 


Exuro89 said:
mgh = 1/2(mg/d)d^2

mgh = 1/2mgd

h = 1/2d

You have to put h instead of d the equation should be mgh = 1/2(mg/d)h^2.
Then your answer will be h = 2d.
 

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