a) Calculate the work done on a 1500-kg elevator car by its cable to l

Given:
${\displaystyle m_{car}=1500kg}$
${\displaystyle \mathrm{△}y=40m}$
${\displaystyle F_s=100N}$
(a)
First of all, we need to find the force of the cable.
Noting that the friction force always working in the opposite direction of the motion of the object.
We know, from Newton's second law, that
${\displaystyle \sum F_y=F_{cable}-mg-F_s=m{a}_y}$
And we know that ${\displaystyle a_y=0}$ since the velocity is constant.
hence,
${\displaystyle F_{cable}-mg-F_s=0}$
${\displaystyle F_{cable}=mg+F_s}$
We know that the work done by some force is given by
${\displaystyle W=Fd\cos \theta }$
so the work done by the cable force on the car during this process is given by
${\displaystyle W_{cable}=F_{cable}\mathrm{△}y{\cos 0}^{\circ }}$
Plug from (1), and noting that cos ${\displaystyle 0^{\circ }=1.0}$
${\displaystyle W_{cable}=(mg+F_s)\mathrm{△}y}$
Now plug the given;
${\displaystyle W_{cable}=[\left(1500x9.8\right)+100]\times 40}$
${\displaystyle W_{cable}=5.92x{10}^{\circ }J}$
(b)
The work done by the gravitational force is given by
${\displaystyle W_{cable}=F_{g}d\cos \theta }$
${\displaystyle W_{cable}=mg\mathrm{△}y\cos \theta }$
Noting that \theta= 180° since the angle between the gravitational force and the velocity is ${\displaystyle {180}^{\circ }}$
Hence,
${\displaystyle W_{gravity}=mg\mathrm{△}y{\cos 180}^{\circ }}$
Noting that cos ${\displaystyle {180}^{\circ }=-1}$
${\displaystyle W_{gravity}=mg\mathrm{△}y}$
Plug the given;
${\displaystyle W_{gravity}=-1500\times 9.8\times 40}$
${\displaystyle W_{gravity}=-5.88\times {10}^{5}J}$
(c)
The total work done on the lift is the work done by the net force exerted on the car during this process.
Hence,
${\displaystyle W_{tot}=F_{net}\theta y}$
We know that the net force exerted on the car during this process is zero since the car is moving at a constant speed (see part a above).
So,
${\displaystyle W_{tot}=0\times 40}$
${\displaystyle W_{tot}=0J}$

Given
mass of elevator m=1500 kg
Lifting height =40 m
Friction Force F=100 N
The cable has to work against gravitational force and friction Force
Increase in Potential Energy of Elevator =mgh=1500$\times$9.8$\times$40=588 kJ
thus work done by gravitational force = 588 kJ
Work done against friction force = F * x =100 * 40=4 kJ
Total Work done by the cable = 592 kJ