Answer:
62,400 J
Step-by-step explanation:
As the problem statement tells you, the work done to lift the mass is the product of the force applied and the distance the mass is lifted. Here, the force decreases as the height of the mass increases.
Setup
For x meters of exposed cable, the force required to lift it is 3x newtons. The force required to lift the object is 1500 newtons, so the total force required to lift the masses an incremental distance (dx) is (3x +1500) newtons.
The work done lifting the mass that distance is the product of force and distance. The total work done is the integral of the incremental work over the total of distance lifted:
[tex]\displaystyle W = \int_0^{40}(3x+1500)\,dx[/tex]
Solution
The value of the integral is ...
[tex]\displaystyle W=\left.\left(\dfrac{3}{2}x^2+1500x\right)\right|_{x=0}^{x=40}=40\left(\dfrac{1}{2}(3\cdot40)+1500\right)=62\,400\qquad\text{joules}[/tex]
The work done to lift the cable and weight is 62,400 joules.
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Additional comment
One reason for writing the integral evaluation this way is to demonstrate that the total of work done is that required to lift the weight (1500 N, 40 m) and that required to lift the center of mass of the cable (120 N, 20 m).
This will generally be the case for "disappearing mass" or "increasing distance" problems. Pumping from a pool is a similar kind of problem with a similar kind of solution.
Effectively, you're doing work on the center of mass of the system, moving it from its initial position to the top of the building.
