Calculating Spring Constant from Conservation of Energy Equation

Click For Summary
To calculate the spring constant using the conservation of energy, the initial and final states must be analyzed with the equation that includes kinetic energy, gravitational potential energy, and elastic potential energy. The initial velocity is 3.0 m/s, and the final velocity is 1.5 m/s, with height changes of -0.1 m and -0.25 m, respectively. The displacement of the spring under static load is determined by the mass and spring constant, leading to the equations Xo = mg/K and Xf = mg/K - ΔX, where ΔX is the change in height. The user suspects an algebraic error in their calculations and seeks guidance to resolve the issue. Properly applying these principles should yield the correct spring constant.
Ajeezy
Messages
6
Reaction score
0

Homework Statement



A mass 1 kg mass is hung from a spring.
Calculate the spring constant from the given data. Velocity initial= 3.0 m/s Velocity final= 1.5 m/s. The initial height traveled is 0.1 meters and the final is 0.25 meters. These height values will be negative since i picked my starting point at the origin so they have moved below the X making them negative.



Homework Equations


I know i am supposed to use the conservation of energy equation with the potential elastic factored into the total PE. the equation would be (1/2*m*v^2)+(mgh)+1/2*k*x^2)=(1/2*m*v^2)+(mgh)+1/2*k*x^2) where the left is the initial and the right is the final. However i keep getting the wrong answer. I think I am doing a simple algebraic error in my calculations. any guidance is appreciated


The Attempt at a Solution

 
Physics news on Phys.org
I think you have to determine the initial displacement of the spring under the static load:

Xo = mg / K

Then final displacement is:

Xf = mg / K - \DeltaX

Where \DeltaX = 0.25m - 0.1m = 0.15m

Put that in your equation and do the transformations and it should give you the correct answer.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
29
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 17 ·
Replies
17
Views
1K
  • · Replies 14 ·
Replies
14
Views
4K
Replies
7
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
17
Views
2K