Work energy question. Rock down a hill.

Click For Summary
SUMMARY

The discussion focuses on a physics problem involving a fish attached to a vertical spring and its behavior when lowered to equilibrium versus when dropped. The key equations utilized include the Work Energy Theorem and Hooke's Law, represented as F = -kx. The correct derivation shows that when the fish is dropped, the maximum stretch of the spring is twice the initial stretch distance (h = 2d), correcting earlier misconceptions about the relationship between gravitational potential energy and spring potential energy.

PREREQUISITES
  • Understanding of the Work Energy Theorem
  • Familiarity with Hooke's Law (F = -kx)
  • Basic knowledge of gravitational potential energy (PE = mgh)
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the derivation of the Work Energy Theorem in detail
  • Learn about the applications of Hooke's Law in real-world scenarios
  • Explore the concept of energy conservation in mechanical systems
  • Investigate the dynamics of oscillating systems involving springs and masses
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of energy conservation and spring dynamics.

Exuro89
Messages
33
Reaction score
1
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:
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 58 ·
2
Replies
58
Views
3K
Replies
7
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
17
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K