JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Work, Power & Energy

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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})

(A) $1.5 ms^{- 1}$
(B) $2.0 ms^{- 1}$
(C) $1.7 m / s^{- 1}$
(D) $1.9 m / s^{- 1}$ ✓ Correct

Answer: (D)

Solution

At full load the motor supplies enough force to carry the weight and overcome friction, with

P = (m g + F_{fric}) v

. Here

P = 60 \times 746 = 44760 W

and

m g + F_{fric} = 2000 \times 10 + 4000 = 24000 N

, so the steady speed is

v = \frac{P}{m g + F _{fric}} = \frac{44760}{24000} \approx 1.9 m/s

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Single correct option

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