Solving Conservation of Energy Problems

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SUMMARY

This discussion focuses on solving conservation of energy problems in physics, specifically using the equation Total Energy = Kinetic Energy + Potential Energy. The participants clarify the application of the formula T.E = 1/2mv^2 + mgh, emphasizing the importance of correctly denoting variables and simplifying terms. Key insights include the necessity of maintaining the gravitational constant 'g' and the correct manipulation of kinetic energy terms. The final expression derived is v = sqrt(v^2 + gh), which accurately represents the relationship between kinetic and potential energy.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Familiarity with kinetic and potential energy equations
  • Basic algebraic manipulation skills
  • Knowledge of gravitational acceleration (g = 9.8 m/s²)
NEXT STEPS
  • Study the derivation of energy conservation equations in physics
  • Learn how to apply the conservation of energy principle to different scenarios
  • Explore advanced topics in mechanics, such as energy transformations
  • Practice solving problems involving kinetic and potential energy with varying heights
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Students studying physics, educators teaching energy concepts, and anyone looking to improve their problem-solving skills in mechanics.

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.
 
Last edited:

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