Energy Efficiency: Input & Output (Work/Energy)

AI Thread Summary
To determine the input energy provided by a 54 kg athlete running at 11 m/s with an efficiency of 85%, the kinetic energy formula Ek = 1/2 mv^2 is used to find the output energy. The output energy is calculated based on the athlete's speed, and the efficiency formula Eout/Ein x 100% is applied to relate input and output energy. The confusion regarding the efficiency percentage arises from whether it is expressed as a decimal (0.85) or a percentage (85), which affects the equation's structure. Understanding that 85% equals 0.85 clarifies how to correctly apply the efficiency in calculations. This discussion emphasizes the importance of correctly interpreting efficiency in energy calculations.
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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