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    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

    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

    Byju's Answer Standard XI Physics

    Power, Force and Velocity

    An electric l... Question

    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 ) A 34500 B 23000 C 20000 D 23500 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 ⋅ V = 23000 × 1.5 P = 34500 W Suggest Corrections 19

    SIMILAR QUESTIONS

    Q. An elevator in a building can carry a maximum of

    10

    persons, with the average mass of each person being

    68 kg

    . The mass of the elevator itself is

    920 kg

    and it moves with a constant speed of

    3 ms − 1

    . The frictional force opposing the motion is

    6000 N

    . If the elevator is moving up with its full capacity, the power delivered by the motor to the elevator

    ( g = 10 ms − 2 ) must be at least

    Q. 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 )

    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.

    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 .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?

    View More

    स्रोत : byjus.com

    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)

    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)

    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)

    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)

    A 20000 B 34500 C 23500 D 23000

    Solution:

    P=(mg+f)v =(2000×10+3000)1.5 =23000×1.5 =34500 N

    P=(mg+f)v=(2000×10+3000)1.5=23000×1.5=34500 N

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    स्रोत : infinitylearn.com

    [Solved] An electric lift with a maximum load of 2000 kg (lift + pass

    CONCEPT: The power is defined as the product of the tension applied in the electric lift to the velocity and it is written as; P = Tv    ----(1) He

    Home Physics Laws of Motion Common forces in mechanics Friction

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    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)

    23500 23000 20000 34500

    Answer (Detailed Solution Below)

    Option 4 : 34500 Crack with

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    Detailed Solution

    Download Solution PDF

    CONCEPT:

    The power is defined as the product of the tension applied in the electric lift to the velocity and it is written as;

    P = Tv    ----(1)

    Here, P is the power, T is the tension and v is the velocity.

    and weight is defined as the product of mass and acceleration due to gravity and it is written as;

    W = mg

    Here 'm' is the mass and 'g' is the acceleration due to gravity.

    CALCULATION:Given : v = 1.5 m/s

    m = 2000 kg F = 3000 N

    Here we are given that the velocity is constant, and acceleration is defined as the rate of change of velocity with respect to time, therefore for the constant velocity the acceleration (a) is zero.

    a = 0

    The figure of an electric lift is shown below;

    Now, as we see in the figure the force acting downwards, therefore,

    The tension T = W + F

    ⇒ T = mg + F ⇒ T = 2000 × 10 + 3000 ⇒ T = 20000 + 3000 ⇒ T = 23000 N

    Now, using equation (1) we have;

    P = Tv ⇒ P = 23000 × 1.5

    ⇒ P = 34500 watts.Hence, option 4) is the correct answer.

    Download Solution PDF

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    स्रोत : testbook.com

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