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:|

i will watch. good morning sensei

Gmornin :)

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.

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?

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?

Half the length. Yes, Amistre.

weight of cable being 4 lb/ft

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.

:\[W = \int\limits_{0}^{150}4x dx = 45,000.\]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 ; ^)