Energy Efficiency: Input & Output (Work/Energy)

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Homework Help Overview

The problem involves calculating the input energy provided by an athlete running at a certain speed, given their mass and efficiency. The subject area pertains to work and energy principles, particularly focusing on kinetic energy and efficiency calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between input and output energy, questioning how to determine the output energy without height information. The use of kinetic energy formulas is suggested, and there is a clarification regarding the application of efficiency in calculations.

Discussion Status

Participants are actively engaging with the problem, exploring different approaches to find the output energy. Some guidance has been provided regarding the use of kinetic energy and the interpretation of efficiency in the context of the problem.

Contextual Notes

There is a noted absence of height in the problem setup, which raises questions about the appropriate energy calculations. The discussion also touches on the proper use of efficiency percentages in equations.

harujina
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Homework Statement



In a race, a 54 kg athlete runs from rest to a speed of 11m/s on a flat surface. The athlete's body has an efficiency of 85% during the run. How much input energy did the athlete provide?

Homework Equations



efficiency = E out/E in x 100%

The Attempt at a Solution



I know E in = E out/efficiency
and efficiency = 85%, m = 54 kg, vi = 0, vf = 11 m/s
but how can I find E out?
Eg = mgh ?
There's no height though...?
 
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harujina said:

The Attempt at a Solution



I know E in = E out/efficiency
and efficiency = 85%, m = 54 kg, vi = 0, vf = 11 m/s
but how can I find E out?
Eg = mgh ?
There's no height though...?

His energy output is entirely kinetic! No need for a height.
 
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rock.freak667 said:
His energy output is entirely kinetic! No need for a height.

Oh, right.
Could I use Ek = 1/2mv^2 to solve this then?
 
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harujina said:
Oh, right.
Could I use Ek = 1/2mv^2 to solve this then?

Yes you can :approve:
 
The normal formula of efficiency is: Eout/Ein x 100% = Eff%. Why is it that when solving this question you guys got rid of the "x100%" ?
 
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Matthew L said:
The normal formula of efficiency is: Eout/Ein x 100% = Eff%. Why is it that when solving this question you guys got rid of the "x100%" ?
Good question regarding the basics. The answer lies in how we ll put efficiency into this equation. IF for example efficiency is 85% and we put it in the equation the efficiency as just 85 then we got to keep the 100. If we put efficiency as 0.85 then we got to remove the 100.

The symbol % in here just means division by 100. When we say efficiency is 85% we actually mean it is 85/100=0.85. When we multiply x100% we multiply by 100 and then divide by 100, so its like multiplying by 1.
 
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Delta2 said:
Good question regarding the basics. The answer lies in how we ll put efficiency into this equation. IF for example efficiency is 85% and we put it in the equation the efficiency as just 85 then we got to keep the 100. If we put efficiency as 0.85 then we got to remove the 100.

The symbol % in here just means division by 100. When we say efficiency is 85% we actually mean it is 85/100=0.85. When we multiply x100% we multiply by 100 and then divide by 100, so its like multiplying by 1.
thank you!
 
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