# 25 lbs 12 5 lbs the work done in moving the bottom

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·25 lbs.=12.5 lbs. , the work done in moving the bottom half of the chain5 ft. upward isW1=(12.5 lbs. )·(5 ft. )=62.5 ft.-lbs.25
It remains to examine the second14of the chain. Here, each piece of the chain islifted a distance equal to twice its original distance from the pointPlocated14of theway down the chain (eg., the pointPdoes not move at all, but the midpoint of the chainstarts 2.5 ft. belowP, and ends up 2.5 ft. aboveP, which is 2·2.5 ft.=5 ft. aboveits original position). Therefore, if we setxto be the starting distance belowP, thenwe letxrange over 0x2.5 ft., and each piece of thicknessdxis moved upward adistance 2x. The weight density of the chain is given by25 lbs.10 ft.=2.5 lbs./ft.This means that a piece of chain of lengthdxwill have weight 2.5dxlbs., and since itis moved a distance 2x, the work done is given by the integralW2=integraldisplay2.50(2.5dx)(2x)=integraldisplay2.505xdx=2.5x2|x=2.5x=0=2.5·(2.5)20=15.625 ft.-lbs.Putting the two parts of the work together, the total work done isW1+W2=62.5 ft.-lbs.+15.625 ft.-lbs.=78.125 ft.-lbs.26
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