Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 100 n. (b) what is the work done on the lift by the gravitational force in this process? (c) what is the total work done on the lift?

The weight of the elevator is

W = (1500 kg) * (9.8 m/s²) = 1.47 x 10⁴ N

Because the frictional resistance is R = 100 N, the tension in the cable for dynamic equilibrium is

T = W + R = 1.47 x 10⁴ + 100 = 1.48 x 10⁴ N

By definition, the work done in lifting the elevator by 40 m is

(1.48 x 10⁴ N) * (40 m) = 5.92 x 10⁵ J = 592 kJ

Answer: 592 kJ (or 5.92 x 10⁵ J)