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

- amistre64

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

- jamiebookeater

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- anonymous

:|

- anonymous

i will watch. good morning sensei

- amistre64

Gmornin :)

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## More answers

- amistre64

im sure we are starting at one end and winding the whole thing up, at least that is what I pictured

- anonymous

honestly if i pretended to help i would be making things up. never had a physics class in my life.

- amistre64

(150-x) would be the length of each "portion" which then weighs 4 lb/ft
maybe:
\[\int_{0}^{150}(150-x)4\ dx\]

- anonymous

its very easy

- amistre64

im sure it is, but im wondering if I did the integral correctly ...

- anonymous

i will learn something...

- amistre64

i got another one about fluid density pressure after this :)

- anonymous

physics class?

- anonymous

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.

- amistre64

calculus1

- anonymous

It assumes it moves vertically, just enough to leave the ground.

- anonymous

Then the vertical distance is 75 ft.

- anonymous

So W = (600lb)(75ft) = 45,000 ft. lb.

- amistre64

so the 75 comes from 150/2 right?

- amistre64

how would we set that up as an integral?

- anonymous

I picture one end being lifted up until the other end is just about to leave the ground. Is that the rright scenario?

- anonymous

Half the length. Yes, Amistre.

- amistre64

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?

- amistre64

weight of cable being 4 lb/ft

- anonymous

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.

- anonymous

If the weight is 4 lb/ft, then x ft weighs (4x) lb, so 4x is the integrand.

- anonymous

The integral evaluatex to 2x^2, where x = 150. Works out to 2* 22500 =
45,000 ft. lb.

- amistre64

so if anything, I would have been off by makeing the force = length*weight?
[150ft - (x ft off the ground) ] * 4 lb/ft?
\[\int_{0}^{150}(150-x)4\ dx\]
\[\int_{0}^{150}(500-4x)\ dx\implies500x-2x^2|^{150}\]

- anonymous

:\[W = \int\limits_{0}^{150}4x dx = 45,000.\]The length is x ft, and the weight pre foot is 4 lb/ft.

- amistre64

thanx :)

- anonymous

Remember the definition of work?

- anonymous

:^)

- amistre64

Work = Force x Distance yes :)

- anonymous

That's the idea ; ^)

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