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JEE Main20207 Jan 2020Morning ShiftPhysicsLaws of MotionActual

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

Options

  1. A1.7   m   s - 1
  2. B1.9   m   s - 1
  3. C1.5   m   s - 1
  4. D2.0   m   s - 1

Correct answer

B. 1.9   m   s - 1

Step-by-step solution

4000 × V + m g × V = P 60 × 746 4000 + 20000 = V V = 1.9 m s - 1

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