Total Mechanical Energy: Mass or No Mass?

Click For Summary

Homework Help Overview

The discussion revolves around the concept of total mechanical energy, specifically addressing the inclusion or exclusion of mass in the energy equations. The original poster questions why the solutions provided do not include mass when calculating total mechanical energy, which they initially believed to be expressed as mgh + 0.5mv².

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the idea of energy per unit mass and how it relates to the total mechanical energy equation. They discuss the implications of dividing the energy by mass and the usefulness of this approach in the context of the problem.

Discussion Status

Some participants have provided clarification regarding the concept of energy per unit mass, suggesting that the problem is framed in such a way that it is more relevant to consider energy without explicitly including mass. There is an acknowledgment of the reasoning behind this approach, but no explicit consensus has been reached on the broader implications.

Contextual Notes

The discussion hints at the possibility that the problem is designed to focus on energy per unit mass, which may influence how participants interpret the equations involved. There is also mention of practical considerations regarding efficiency in real-world applications, such as hydro plants.

TyErd
Messages
297
Reaction score
0

Homework Statement



In the diagram I have attached the initial question is to find the total mechanical energy. however in the solutions which I have placed under the diagram, they don't use the mass (m). So total mechanical energy i thought was = mgh + 0.5m[itex]v^{2}[/itex] but they use gh + 0.5[itex]v^{2}[/itex] . no mass. why

i think its something to do with the question asking to be in per unit mass but i don't see why.

Homework Equations


The Attempt at a Solution

 

Attachments

  • dadadadad.png
    dadadadad.png
    23.7 KB · Views: 448
Physics news on Phys.org
They ask you for the energy per unit mass that is present, and so they divide both sides of the equation by the mass. It's nothing more complicated than that. It's a more useful quantity to consider in this situation than the amount of energy present, which depends upon how big a "parcel" of water you consider.
 
Think of it this way: every kilogram of water that passes by carries with it 0.887 kJ of mechanical energy. You have the volume flow rate (which is the volume of water that flows by in 1 second). You can use that to determine the mass flow rate (the rate at which water flows by in kg/s). Combine that with your energy per unit mass, and suddenly you have the energy per second being generated (EDIT: or at least the energy per second that could be generated if all of that mechanical energy in the water flow could be transferred to some other system with 100% efficiency -- in a real hydro plant the efficiency is sure to be lower).
 
in essence,

per unit mass = mass = 1

so mgh = 1gh = gh

same with KE
 
thankyou guys. helps a lot
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K