Finding work for lifting a chain with winch on top of a building

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SUMMARY

The discussion focuses on calculating the work done in reeling up a 60 m chain using a winch positioned 70 m above ground. The relevant formula for work, W = F * D, is applied, where F is derived from the mass per unit length (μ = 1.2 kg/m) and gravitational acceleration (g = 9.81 m/s²). The user attempts to integrate the force over the distance but arrives at an incorrect result of 28252.8 Joules instead of the expected 42200 Joules. The confusion arises from the proper application of the chain's length in the calculations.

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Homework Statement



A winch is positioned on top of a building, a distance 70 m above ground level. A chain of length 95 m and a mass per unit length of 1.2 kg/m hangs from the winch along the side of the building. Find the work done (in Joules) in reeling up 60 m of the chain.

Homework Equations



W=FxD

F=μLg (according to a few examples)

The Attempt at a Solution



hmm i followed a few examples seeing if it can help me get to the right answer
ok so i know that
dW=F*dy F=μLg
so μ=1.2, g=9.81, L=(70-y) ...got that from the equation but not too sure about L
...if i put that together to solve for W
W=∫ μLg dy from 0 to 60 ...60 because its reeling it up by 60
then i do all the integral magic and i get 28252.8
but the answer is suppose to be 42200 >_<

i used most of the number except for the length of the chain...not too sure what to do with it... i thought i can use to for my limits 95-60 and 95-70 but...i think that's just saying I am only lifting its 25m to 35m =/hope i got the right idea...thanks
 
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this one may be easier to balance work & potential energy (ps i didn't go through all the working)
 

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