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

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The discussion focuses on calculating the input work required to lift an 18kg mass using a 90% efficient lever. The participant attempts to use the equation (Fout/Fin)/(Din/Dout) but expresses confusion over its format, noting the absence of an equals sign. They calculate the input work to be 98.1J, which aligns with another participant's answer. Clarification is provided that the correct interpretation of the equation should be (Fout/Fin) = (Din/Dout). The conversation emphasizes understanding the efficiency of simple machines in relation to work input and output.
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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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