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amistre64
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Find the amount of work done by lifting 150 ft of wire that weighs 4 lbs/foot.
 3 years ago
 3 years ago
amistre64 Group Title
Find the amount of work done by lifting 150 ft of wire that weighs 4 lbs/foot.
 3 years ago
 3 years ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.0
i will watch. good morning sensei
 3 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
Gmornin :)
 3 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
im sure we are starting at one end and winding the whole thing up, at least that is what I pictured
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
honestly if i pretended to help i would be making things up. never had a physics class in my life.
 3 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
(150x) would be the length of each "portion" which then weighs 4 lb/ft maybe: \[\int_{0}^{150}(150x)4\ dx\]
 3 years ago

elecengineer Group TitleBest ResponseYou've already chosen the best response.0
its very easy
 3 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
im sure it is, but im wondering if I did the integral correctly ...
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
i will learn something...
 3 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
i got another one about fluid density pressure after this :)
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
physics class?
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
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.
 3 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
calculus1
 3 years ago

elecengineer Group TitleBest ResponseYou've already chosen the best response.0
It assumes it moves vertically, just enough to leave the ground.
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
Then the vertical distance is 75 ft.
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
So W = (600lb)(75ft) = 45,000 ft. lb.
 3 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
so the 75 comes from 150/2 right?
 3 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
how would we set that up as an integral?
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
I picture one end being lifted up until the other end is just about to leave the ground. Is that the rright scenario?
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
Half the length. Yes, Amistre.
 3 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
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?
 3 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
weight of cable being 4 lb/ft
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
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.
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
If the weight is 4 lb/ft, then x ft weighs (4x) lb, so 4x is the integrand.
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
The integral evaluatex to 2x^2, where x = 150. Works out to 2* 22500 = 45,000 ft. lb.
 3 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
so if anything, I would have been off by makeing the force = length*weight? [150ft  (x ft off the ground) ] * 4 lb/ft? \[\int_{0}^{150}(150x)4\ dx\] \[\int_{0}^{150}(5004x)\ dx\implies500x2x^2^{150}\]
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
:\[W = \int\limits_{0}^{150}4x dx = 45,000.\]The length is x ft, and the weight pre foot is 4 lb/ft.
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
Remember the definition of work?
 3 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
Work = Force x Distance yes :)
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
That's the idea ; ^)
 3 years ago
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