Work-energy theorem , not sure what I am doing wrong.

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SUMMARY

The discussion focuses on applying the work-energy theorem to a skater's motion on a frictionless ice rink encountering a rough patch. The skater's speed decreases from 6.0 m/s to 3.48 m/s (42% of the original speed) due to a friction force that is 20% of her weight. The correct application of the work-energy theorem involves calculating the change in kinetic energy (KE) and relating it to the work done by friction. The formula used should be -0.2mg * d = (1/2)m(vf^2 - vi^2), where 'd' is the length of the rough patch.

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On an essentially frictionless, horizontal ice rink, a skater moving at 6.0 m/s encounters a rough patch that reduces her speed by 42 \% due to a friction force that is 20 \% of her weight.

Use the work-energy theorem to find the length of this rough patch.

i do: [.42 * (.5) * (6)^2]/ [.2 * 9.8]=3.85
but i get the wrong answer can anyone help, does the answer need to be negative, what am i doing wrong.
 
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What does the work-energy theorem state?

What is the change of the KE if the speed is reduced to 42% of the original 6 m/s ?

ehild
 
K_2-K_1 where k=1/2*m*v^2, thank you, it would be .58v not .42, in the equation
-.2gs = (1/2)mvf^2 - (1/2)mvi^2 where the m cancel
 

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