Work input/output, efficient probelm, just need someone to check my work

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SUMMARY

The discussion centers on calculating the input work required to lift an 18kg mass using a lever with 90% efficiency. The participant correctly identifies the output force (Fout) as 176.6N and applies the work-energy principle to derive the input work (Fin) as 98.1J. A clarification is made regarding the equation format, emphasizing that the relationship should be expressed as (Fout/Fin) = (Din/Dout) for proper understanding.

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Work input/output, efficient probelm, just need someone to check my work :)

Homework Statement


Because there is very little friction, the lever is an extremely efficient simple machine. Using a 90% efficient lever, what input work is required to lift an 18kg mass through a distance of 0.5m?


Homework Equations



(Fout/Fin)/(Din/Dout)

The Attempt at a Solution



(Fout/Fin)/(Din/Dout)
(176.6N/Fin)/(Din/.5m)=.9

so you get (Fin)(Din)=98.1 which FD is work so 98.1J is the answer correct?
 
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Well, I do not quite understand (Fout/Fin)/(Din/Dout) as being an "equation". There is no = sign!

That said, when I work the problem, my answer matches yours. I think you mean to write (Fout/Fin) = (Din/Dout), which is legitimate!
 

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