Answer :
(a) Work done on the car by its cable will be 5.92 * [tex]10^{5}[/tex] J.
(b) Work done on the car by the gravitational force will be -5.92 * [tex]10^{5}[/tex] J.
(c) Total work done on the car will be 0 J.
Work done is the product of force and displacement.
- Mathematically, Work-done = W = Force * Distance
Mass of the car = m = 1500 kg
Distance lifted by the cable = d = 40 m
Frictional force experienced by the car = Fs = 100 N
(a) According to the Newton's second law of motion
F = Fcable - mg - Fs = ma
Fcable = mg + Fs
Work done by the same force = W1
W1 = Fcable * d
W1 = (mg + Fs) * d
W1 = (1500*9.8 + 100) * 40
W1 = 592600
W1 = 5.92 * [tex]10^{5}[/tex] J
(b) Work done by the gravitational force = W2 = mg * d
W2 = 1500*(-9.8)*40
Since the acceleration is acting downwards 'g' will be considered with a negative sign.
W2 = -5.88 * [tex]10^{5}[/tex] J
(c) The total work done on the lift is the work done by the net force exerted on the car during this process:
W3 = Fnet * d
Fnet = -5.88 * [tex]10^{5}[/tex] J + 5.92 * [tex]10^{5}[/tex] J
Fnet = 0
W3 = 0 * 40
W3 = 0 J
To know more about Work done,
https://brainly.com/question/13741431
#SPJ4
