# 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?

Mar 13, 2015

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 $m \cdot g$ (Newton) and $1 J = 1 N m$ so the work is:
${W}_{w} = m \cdot g \cdot h = 150 \cdot 9.8 \cdot 50 = 73500 J = 73.5 k J$

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:
$F \left(x\right) = 2. x \cdot g = 19.6 x$ (Newton)

What we now need is the integral over $F \left(x\right)$ between $x = 50 \mathmr{and} x = 0$

${W}_{c} = {\int}_{50}^{0} F \left(x\right) \cdot \mathrm{dx} = {\int}_{50}^{0} 19.6 x \cdot \mathrm{dx} =$

${W}_{c} = {|}_{50}^{0} 9.8 {x}^{2} = 24500 J = 24.5 k J$ (!)

Answer : Total Work= ${W}_{w} + {W}_{c} = 73.5 + 24.5 = 98 k J$

(!) 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.