Ayush G. answered 03/05/20
I ensure an effective learning experience for all my students.
Assuming the object was lifted at almost 0 speed ie maintaining equilibrium,
let the length of the cable at any time be x.
In the beginning, x = 50 ft.
In the end, x = 0 ft.
Force due to the object = 30 lb
Force due to the cable at length x = 2x lb
total force at any given point while lifting the object = (30 + 2x) lb
Work done = Integral (F.dx) over the length of the entire distance of 50 ft.
Integrate this and you will get the following results...
Work done = Integral of (30 + 2x)dx
The final work done comes out to be = -30x - x^2. Apply the limits from 50 to 0.
You will receive the work done = - (-1500 - 2500) = 4000 lb.ft