Conservation of energy / work problem

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SUMMARY

The discussion centers on a conservation of energy problem involving a hill with a height difference of 30 meters. Participants clarify the algebraic steps necessary to simplify the equation, specifically focusing on the identification of initial (yi) and final (yf) heights. The correct calculation for the height difference is confirmed as 30m - 0m. Additionally, the importance of factorization in the final algebraic expression is emphasized, particularly the terms -2gyf + 2gyi.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Basic algebraic manipulation skills
  • Familiarity with gravitational potential energy equations
  • Knowledge of height difference calculations in physics
NEXT STEPS
  • Review gravitational potential energy formulas
  • Practice algebraic factorization techniques
  • Explore conservation of energy problems in physics textbooks
  • Learn about energy transformations in mechanical systems
USEFUL FOR

Students studying physics, educators teaching energy concepts, and anyone interested in mastering algebraic problem-solving in the context of energy conservation.

adams_695
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Homework Statement
Call the system the bicycle and the rider. Use the work-energy equation and W = Fd. Assume the cyclist and air do not heat up. The work-energy equation is Kf + Ugf = Ki + Ugi + W.
Relevant Equations
Kinetic energy & gravitational energy
Answer- Physics .jpg
Attempt 1- Physics.jpg
Attempt 2- Physics.jpg

If someone could advise what I've done wrong it would be much appreciated. How have they eliminated the initial and final for y, and simplify only to y? Also, how did they simplify to a positive 2? What algebraic steps have I missed? Thanks for your help.
 
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I would not call that a problem statement. What is the actual question?
 
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haruspex said:
I would not call that a problem statement. What is the actual question?
695672E7-6A77-4F8D-8425-24ED89C62321.png


Sorry here it is. Question 9. Thanks!
 
You are told the top of the hill is 30m higher than the bottom of the hill. So what is yi-yf?
 
haruspex said:
You are told the top of the hill is 30m higher than the bottom of the hill. So what is yi-yf?

30m-0m.

Is this how I went wrong when solving algebraically in the image of my working?
 
There was nothing wrong in what you worked out. As haruspex pointed out, identify yi and yf in your own algebra and you are done.
 
adams_695 said:
30m-0m.

Is this how I went wrong when solving algebraically in the image of my working?
The last line of your working has the terms -2gyf+2gyi. Factorise.
 

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