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QJEEM20Phf37f: JEE Main 2020 Physics - Class 11 JEEM20Phf37f Solution: A 60 HP electric motor lifts an elevator having a maximum total …

A 60 HP 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 ms-2)

Multiple Choice Options

  • A 1.5 ms-1
  • B 1.7 ms-1
  • C 2.0 ms-1
  • D 1.9 ms-1

Step-by-Step Verified Solution



F = mg + f

F = 20000 + 4000 = 24000 N

We know, Power(P) = Fv

⇒\Rightarrow v = PF{P \over F} = 60×74624000{{60 \times 746} \over {24000}}

⇒\Rightarrow v ≈\approx 1.9 m/s

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