A 60 HPelectric motor lifts an elevator with a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to (Given 1 HP = 746 W, g = 10 m/s²)

Calculation: 

P = Fv + Mgv

Where, 

Fv is the power required to overcome the frictional force F while moving at speed v and,

Mgv is represents the power required to lift the elevator and its load against gravity at speed v

Applied power = 4000 × v + 20000 v

⇒ 60 × 746 = 4000 v + 20000 v

⇒ v = 1.865

⇒ v = 1.9 m/s Rounding it off.

∴ The Correct answer is Option (4): 1.9 m/s