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 )

Question

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) 1.7 ms- 1 (C) 2.0 ms- 1 (D) 1.9 ms- 1

Tardigrade Admin · Accepted Answer

4000 Ã V + mg Ã V = P (60×746/4000+20000)=V V=1.86 m/s. ≈ 1.9 m/s.

Q. 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 : ( $1 H P = 746 W, g = 10 m s^{- 2}$ )

$1.5 m s^{- 1}$

$1.7 m s^{- 1}$

$2.0 m s^{- 1}$

$1.9 m s^{- 1}$

$4000 \times V + m g \times V = P$
$\frac{60 \times 746}{4000 + 20000} = V$
$V = 1.86 m / s . \approx 1.9 m / s .$

Q. 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 : (1HP=746W,g=10ms−2 )

1.5ms−1

1.7ms−1

2.0ms−1

1.9ms−1