Elephant in the way kinetic energy problem.

AI Thread Summary
The discussion revolves around solving a kinetic energy problem involving a 3 kg object released from a height of 5 meters. The conservation of mechanical energy principle is applied, equating gravitational potential energy to elastic potential energy. The key formula derived is x = √((2mgh)/k), where k is the spring constant. Substituting the given values leads to an approximate solution of 1.73205 meters for the spring's compression. The explanation clarifies how to approach the problem effectively.
Bob Loblaw
Messages
67
Reaction score
0
"Elephant in the way" kinetic energy problem.

Homework Statement


http:/www.imgred.com/http://static.scribd.com/docs/736egzypod07v_files/image004.jpg

I found this on one of those "OMG FUNNY" internet web pages. I am trying to attempt a solution.

Homework Equations



Mechanical energy is conserved so Einitial=Efinal. Uinitial+Kinitial=Ufinal+Kfinal.
mgyinitial+1/2mvinitial^2=mgyfinal+1/2mvfinal^2

The Attempt at a Solution



KE+PE = Total Energy. When the 3kg object is released from a 5m. I am unsure on how to procede with this problem. How does the force constant k of the spring factor into the solution?
 
Last edited:
Physics news on Phys.org
You're doing it wrong.
How to do this problem:
U(initial) {which is gravitational potential energy} = mgh, where h is 5m, m is 3kg, g is 9.81m/s^2.
U(final) {which is elastic potential energy} = 1/2kx^2, where k is 100.
WE WANT X.
W=delta E
Conservation of energy in sys mass, plane, earth. W=0
0=1/2kx^2 - mgh
mgh=1/2kx^2
(2mgh)/k=x^2
So, x=Square Root ((2mgh)/k)
So, the answer in this problem is SqRt((2(3kg)(9.81m/s^2)(5m))/100)
~SqRt 3 = 1.73205 meters

Hope that helped, if you care at this point, months later.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

Similar threads

Replies
5
Views
2K
Replies
1
Views
2K
Replies
2
Views
3K
Replies
7
Views
6K
Replies
4
Views
1K
Replies
3
Views
1K
Back
Top