Find the amount of work done by lifting 150 ft of wire that weighs 4 lbs/foot.

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Find the amount of work done by lifting 150 ft of wire that weighs 4 lbs/foot.

Mathematics
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i will watch. good morning sensei
Gmornin :)

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im sure we are starting at one end and winding the whole thing up, at least that is what I pictured
honestly if i pretended to help i would be making things up. never had a physics class in my life.
(150-x) would be the length of each "portion" which then weighs 4 lb/ft maybe: \[\int_{0}^{150}(150-x)4\ dx\]
its very easy
im sure it is, but im wondering if I did the integral correctly ...
i will learn something...
i got another one about fluid density pressure after this :)
physics class?
The total weight of the cable is 600 lb. Find the vertical distance the center of mass moves in ft., then multiply that distance by 600, and you have the ansewr in ft. lb. The hard part is finding the vertical distance. The problem, as stated, does not provide enough information on that score.
calculus1
It assumes it moves vertically, just enough to leave the ground.
Then the vertical distance is 75 ft.
So W = (600lb)(75ft) = 45,000 ft. lb.
so the 75 comes from 150/2 right?
how would we set that up as an integral?
I picture one end being lifted up until the other end is just about to leave the ground. Is that the rright scenario?
Half the length. Yes, Amistre.
i dunno if thats the right scenario; i believe the question was asking, my interpretation is this, you have 150 feet of cable suspended by a winch; how much work is required to roll it all up onto the winch?
weight of cable being 4 lb/ft
If you want to set it up as an integral, you can start with\[\int\limits_{0}^{150}F(x) dx,\]where F(x) is the force needed to lift x feet of cable off the ground.
If the weight is 4 lb/ft, then x ft weighs (4x) lb, so 4x is the integrand.
The integral evaluatex to 2x^2, where x = 150. Works out to 2* 22500 = 45,000 ft. lb.
so if anything, I would have been off by makeing the force = length*weight? [150ft - (x ft off the ground) ] * 4 lb/ft? \[\int_{0}^{150}(150-x)4\ dx\] \[\int_{0}^{150}(500-4x)\ dx\implies500x-2x^2|^{150}\]
:\[W = \int\limits_{0}^{150}4x dx = 45,000.\]The length is x ft, and the weight pre foot is 4 lb/ft.
thanx :)
Remember the definition of work?
:^)
Work = Force x Distance yes :)
That's the idea ; ^)

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