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**[SOLVED] Finding Work Done By Friction**

A 1.2g feather is dropped from a height of 1.5m and immediately reaches terminal velocity. If the time to fall is 0.95 s and force of the air resistance on the feather is constant, how much work was done on the feather by friction? Answer: W = -0.016 J

I use the equation W = U + K which turns into W = Uf - Ui + Kf - Ki.

Then Uf + Kf = Ui + Ki + W. I know I have to bring the W to the other side but I don't know why. Does anyone have an explanation?

Then the equation becomes Ui = Kf + W because there's no Ki and Uf.

Then substitute mgh for Ui and 1/2mv^2 for Kf which makes the equation:

W = mgh - 1/2mv^2 but the answer turns out to be positive rather negative. I switch the mgh and 1/2 mv^2 and get a slightly different number it it's right. What am I doing wrong?

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