Work energy question. Rock down a hill.

AI Thread Summary
The discussion revolves around a physics problem involving a fish attached to a vertical spring. The initial question was about the maximum stretch of the spring when the fish is dropped from rest, leading to the application of the Work Energy Theorem. The user derived the spring constant and attempted to set up the energy equations, initially leading to confusion about the relationship between the stretch distance and the maximum distance. After further analysis, it was concluded that the correct maximum stretch of the spring when the fish is dropped is twice the initial stretch distance, resulting in the final equation h = 2d. The conversation highlights the importance of correctly applying energy principles in solving physics problems.
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?
 
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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?
 
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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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