Work energy question. Rock down a hill.

[Exuro89]

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?

 

[Exuro89]

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?

 

[ashishsinghal]

Does animal right activists know about all this?

 

[ashishsinghal]

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.

 

[rwisz]

Your attempt at a solution is on the right track. However, there are a few things to consider. First, the work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. In this case, the fish starts at rest and ends at rest, so the change in kinetic energy is zero. Therefore, the net work done on the fish must also be zero.

Next, you are correct in identifying the energies involved as potential energy of the spring and gravitational potential energy. However, you need to consider the initial and final states of the fish. In the initial state, the fish is at the equilibrium position of the spring and has no gravitational potential energy. In the final state, the fish has been released and falls a distance h, gaining gravitational potential energy. Therefore, the equation should be:

0 = -1/2kd^2 + mgh

where mgh is the final gravitational potential energy gained by the fish.

Now, you correctly identified the force constant as k = (mg)/d. However, in order to find the maximum distance h, you need to solve for it. Rearranging the equation, we get:

h = 1/2kd^2/mg

Substituting in the value for k, we get:

h = (1/2)(mg)/d * d^2/mg

Simplifying, we get:

h = 1/2d

which is the correct answer.

In summary, the maximum distance the fish will fall and stretch the spring is equal to half of the initial stretch distance d. This makes sense intuitively, as the fish will gain gravitational potential energy equal to the potential energy stored in the spring when it is stretched to distance d.

I hope this helps clarify your understanding. Keep up the good work!