Question

How much work is required to lift the load of the way up the shaft if a cable that weighs kg/meter is lifting a load of 150 kg that is initially at the bottom of a 50 meter shaft?

Answer 1

Since you provided no number, let's say the cable is 2 kg/m. You can fill in your own number in the second equation below.

There are two portions to this:

Lifting the weight:
Force needed is `m*g` (Newton) and `1J=1Nm` so the work is:
`W_w=m*g*h=150*9.8*50=73500J=73.5kJ`

Lifting the cable:
While lifting the cable, less and less has to be lifted, so the force needed to do this diminishes. If we call the current length of the cable `x`, then the force needed:
`F(x)=2.x*g=19.6x` (Newton)

What we now need is the integral over `F(x)` between `x=50 and x=0`

`W_c=int_50^0 F(x)*dx=int_50^0 19.6x*dx=`

`W_c= |_50^0 9.8x^2=24500J=24.5kJ` (!)

Answer : Total Work= `W_w+W_c=73.5+24.5=98kJ`

(!) For the calculus-purists: Yes, I embezzled a `-`sign there.
I should have made `x=` the way up from the bottom of the shaft, but that would make the equations more difficult to grasp. And we all know that Work in this case is positive.