Conservation of energy- setting it up

AI Thread Summary
The discussion revolves around confusion regarding the conservation of energy in a physics problem involving helium atoms and energy release. The original poster questions how the initial equation omits potential energy (PE) and whether the kinetic energy (KE) of the helium atoms equals the released energy. They speculate that the energy released (92.0 keV) contributes to the helium particles' KE, suggesting that potential energy is negligible. The teacher confirms that gravitational potential energy is small enough to be disregarded in this context. The poster expresses frustration over the lack of problem-solving examples in class after several days.
Fellowroot
Messages
92
Reaction score
0

Homework Statement



attachment.php?attachmentid=36409&stc=1&d=1307931977.jpg

Homework Equations


The Attempt at a Solution



I do not understand how in the solution they come up with the first equation that I boxed in with green.

I know that the total energy before is equal to the total energy after but I don't understand the logic of how they formed that first equation. It looks like KE+KE=E to me. What happened to the PE potential energy? And how can the KE of the 2 He atoms be equal to the energy released with the atoms?

I would only think that it would be (PE)Beryllium= 2(KE+PE)Helium+92.0keV

Also, I am two days into the class and my physics teacher has NOT solved a single problem. Has this happened to anyone else?

EDIT: OK, I think I got it? 92.0keV of energy is released which goes into the two He particles and somehow they are traveling at Vmax so there is no PE? That's my guess.
 

Attachments

  • physics forms.jpg
    physics forms.jpg
    37.9 KB · Views: 532
Last edited:
Physics news on Phys.org
OK, I asked my teacher why the PE was not in the equation.

He said something about how the gravitational PE is so small that you can neglect it.

This is now the third day of class and he has NOT solved a single problem or done an example on the board.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top