Gravitational potential energy homework

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SUMMARY

The climber's gravitational potential energy increased from 14,000J to 21,000J, resulting in a change of 7,000J. The climber expended 18,000J of energy, leading to an efficiency calculation of approximately 39%. However, the answer key indicates an efficiency of 61%, suggesting a discrepancy. The climber's efficiency is calculated using the formula % Efficiency = (Useful Energy Output / Total Energy Input) x 100%.

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Erwin Schrodinger
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A climber's gravitational potential energy increases from 14000J to 21000J while climbing a cliff. She expends 18000J of energy during this activity. What is the efficiency of this process?

A) 3%
B) 39%
C) 61%
D) 97%


Here is what I did.

Wpg = ∆Epg = Epg2-Epg1 = 21000-14000 = 7000J

% Eff = 100%(useful out)/total in = (100%)7000/18000 ≈ 39%


The answer key says it is 61% though. Is the answer key wrong or did I miss something?
 
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I'd agree with your answer. They key must be wrong.
 

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