Question

Calculate the work done (in J) on a 1200 kg elevator car by its cable to lift it 39.5 m at constant speed, assuming friction averages 145 N.
What is the work done (in J) on the lift by the gravitational force in this process?
What is the total work done (in J) on the lift?

Answer 1

Given m = 1200 kg, h = 39.5 m, friction f = 145 N, g = 9.80 m/s^2.

1) Work done by the cable:
T = mg + f, so W_cable = T h = (mg + f) h
= (1200·9.80 + 145)·39.5 = 11905·39.5 = 4.7025×10^5 J ≈ 4.70×10^5 J.

2) Work done by gravity:
W_gravity = −mgh = −(1200·9.80·39.5) = −4.6452×10^5 J ≈ −4.65×10^5 J.

3) Total work on the lift:
W_total = W_cable + W_gravity + W_friction.
W_friction = −f h = −145·39.5 = −5.7275×10^3 J,
so W_total = 4.7025×10^5 − 4.6452×10^5 − 5.7275×10^3 = 0 J (as expected for constant speed).