A 60 H P electric motor lifts an elevator having a maximum total load capacity of 2000   k g . If the…

A 60HP electric motor lifts an elevator having 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 : 1 HP=746 W, g=10 m s-2
  1. 1.7 m s-1
  2. 1.9 m s-1
  3. 1.5 m s-1
  4. 2.0 m s-1

Solution

4000×V+mg×V=P
60×7464000+20000=V
V=1.9 m s-1

Asked in: JEE Main 2020 (07 Jan Shift 1)

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