An electric lift with a maximum load of $2000 k g$ (lift + passengers) is moving up with a constant speed of $1.5 {m s}^{- 1}$ . The frictional force opposing the motion is $3000 N$ . The minimum power delivered by the motor to the lift in watts is: $(g = 10 {m s}^{- 2})$

34500

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

23000

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20000

No worries! We‘ve got your back. Try BYJU‘S free classes today!

23500

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $34500$
Total force required to move with constant speed,

Force $=$ weight $+$ friction

$F = 20000 + 3000$

$F = 23000 N$

Power, $P = F \cdot V$

$= 23000 \times 1.5$

$P = 34500 W$

Suggest Corrections

299

Similar questions

Q. An elevator can carry a maximum load of

1800 k g (e l e v a t o r + p a s s e n g e r s)

is moving up with a constant speed of

2 m s^{- 1}

. The frictional force opposing the motion is

4000 N

. What is minimum power delivered by the motor to the elevator?

Q. An elevator can carry a maximum load of 1800 kg (elevator + passengers ) is moving up with a constant speed of 2 m/s. The frictional force opposing the motion is 400 N. Determine the minimum power delivered by

Q. An elevator which can carry a maximum load of 1800kg (elevator+passengers)is moving up with a constant speed of 2m/s. The frictional force opposing the motion is 4000N. Determine the minimum power delivered by the motor to the elevator in watt and horsepower .

An elevator of total mass (elevator + passenger) $1800 k g$ is moving up with a constant speed of $2 m s^{- 1}$ . A frictional force of $2000 N$ is opposing its motion. The minimum power delivered by the motor to the elevator is [Take $g = 10 m s^{- 2}$ ]

Q. Sumesh went to Big Bazaar to purchase certain goods. There he has noticed an old lady struggling to carry the goods from one floor to the next. Then Sumesh took her to the lift and showed her how to operate it. The old lady was very happy.

a) What values Sumesh posses?
b) An elevator can carry a maximum load of 1800 kg is moving up with a constant speed of 2 m/s, the frictional force opposing the motion is 4000 N. Determine the minimum power delivered by motor to the elevator in Watts.

Join BYJU'S Learning Program

An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms−1. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is: (g=10 ms−2)

The correct option is A 34500Total force required to move with constant speed, Force = weight + friction F=20000+3000 F=23000 N Power, P=F⋅V =23000×1.5 P=34500 W

An electric lift with a maximum load of $2000 k g$ (lift + passengers) is moving up with a constant speed of $1.5 {m s}^{- 1}$ . The frictional force opposing the motion is $3000 N$ . The minimum power delivered by the motor to the lift in watts is: $(g = 10 {m s}^{- 2})$

The correct option is A $34500$
Total force required to move with constant speed,

Force $=$ weight $+$ friction

$F = 20000 + 3000$

$F = 23000 N$

Power, $P = F \cdot V$

$= 23000 \times 1.5$

$P = 34500 W$