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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i will watch. good morning sensei

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

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

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

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i will learn something...

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It assumes it moves vertically, just enough to leave the ground.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then the vertical distance is 75 ft.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So W = (600lb)(75ft) = 45,000 ft. lb.

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

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I picture one end being lifted up until the other end is just about to leave the ground. Is that the rright scenario?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Half the length. Yes, Amistre.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i 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?

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If the weight is 4 lb/ft, then x ft weighs (4x) lb, so 4x is the integrand.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The integral evaluatex to 2x^2, where x = 150. Works out to 2* 22500 = 45,000 ft. lb.

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

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Remember the definition of work?

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