Solving Conservation of Energy Problems

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Homework Help Overview

The discussion revolves around a conservation of energy problem, specifically involving kinetic and potential energy equations. Participants are attempting to analyze energy transformations at different heights and speeds of an object.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the energy equations, with some noting the need to differentiate between speeds at different heights. There are attempts to simplify the equations and clarify terms, with questions about algebraic manipulation and assumptions regarding energy conservation.

Discussion Status

The discussion is active, with participants providing guidance on algebraic steps and clarifying the use of variables. There is a mix of interpretations regarding the energy terms and their simplifications, but no explicit consensus has been reached on the final form of the equations.

Contextual Notes

Some participants express uncertainty about the information needed to fully solve the problem, indicating potential constraints in the problem setup or missing details.

xRadio
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Homework Equations



Total Energy = Kinetic Energy + Potential Energy
T.E = 1/2mv^2 + m(g)(h)

The Attempt at a Solution



T.E at top
= 1/2mv^2 + m(9.8)(h)

T.E at bottom
1/2mv^2 + m(9.8)(h) = 1/2mv^2 + m(9.8)(h/2)

Please help I have no clue what to do
 
Last edited:
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1/2mv^2 + m(9.8)(h) = 1/2mv^2 + m(9.8)(h/2)[\quote]
The speeds of the object are different, so may want to denote the second speed as v2
 
Oh yeah, I had that on my paper, but forgot to type it in.
 
You look to be on the right track. You just need to solve for v2. Are you having trouble with the algebra? Put "g" back in for the 9.8, to make it clearer.
You can also cancel out m.
 
Last edited:
1/2mv^2 + m(g)(h) = 1/2m[v]^2 + m(g)(h/2)
v^2 + (g)(h) - (g)(h/2) = [v]^2

Is this right? =/ I really suck at simplifying
 
The first line is good. You can simply your gh terms. If you ignore the gh for a moment, you will see that it is really 1-1/2, which is 1/2.

You have also dropped the 1/2 from the kinetic energy terms again.
 
1/2mv^2 + 1/2 = 1/2m[v]^2
v^2 + 1 = [v]^2 *multiply both sides by 2 to eliminate 1/2?
v + 1 = [v]
 
well the help so far has been on the right track.

you can work this out if you choose, but any increase in energy will come from the additional drop of 1/2y*mg

so 1/2mv'^2=1/2mv^2 +1/2 mgh,
so factoring, gives v'=sqrt(v^2+gh)
may be wrong but don't think you have enough information to solve for h.
 
Okay, I got to ask, Bambi's carcass??!
 
  • #10
xRadio said:
1/2mv^2 + 1/2 = 1/2m[v]^2
v^2 + 1 = [v]^2 *multiply both sides by 2 to eliminate 1/2?
v + 1 = [v]


Oh sorry I should have been clearer...you still need the gh in there. I was just pointing out that gh - (1/2)gh = (1/2)gh. So put a gh in where your 1 is.

Also v must remain as v^2, since you have

v^2 + gh = [v]^2

when you take the square root of both sides, you get

(v^2 + gh)^(1/2) = [v]

You can't take the square root of individual terms like you did. You must take it of the entire left side.
 
  • #11
robb_ said:
Okay, I got to ask, Bambi's carcass??!

Robb, I had to laugh as well, I think there is now an industry to make physics problems more entertaining to youth of america. so they dress up the same old dreary problems in ever increasingly bizarre clothes.
 
  • #12
robb_ said:
Okay, I got to ask, Bambi's carcass??!

lol I didn't want to know.
 
  • #13
yess thank you I did it again and got this answer, thank you both so much for all your help.
 
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