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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)
- a)23000
- b)20000
- c)34500
- d)23500
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
An electric lift with a maximum load of 2000 kg (lift + passengers) is...
Fup = 2000 g + 3000
= 23000 N
Minimum power
Pmin = Fv = 23000 × 3 / 2
= 34500 W
Most Upvoted Answer
An electric lift with a maximum load of 2000 kg (lift + passengers) is...
Given Data:
- Maximum load: 2000 kg
- Speed of the lift: 1.5 m/s
- Frictional force: 3000 N
- Acceleration due to gravity: 10 m/s^2
Calculating the Power:
- The force required to lift the load is the sum of the force due to weight and the frictional force.
- Force due to weight = 2000 kg * 10 m/s^2 = 20000 N
- Total force = 20000 N + 3000 N = 23000 N
- Power = Force * Velocity = 23000 N * 1.5 m/s = 34500 W
Therefore, the minimum power delivered by the motor to the lift is 34500 watts, which corresponds to option 'C'.
- Maximum load: 2000 kg
- Speed of the lift: 1.5 m/s
- Frictional force: 3000 N
- Acceleration due to gravity: 10 m/s^2
Calculating the Power:
- The force required to lift the load is the sum of the force due to weight and the frictional force.
- Force due to weight = 2000 kg * 10 m/s^2 = 20000 N
- Total force = 20000 N + 3000 N = 23000 N
- Power = Force * Velocity = 23000 N * 1.5 m/s = 34500 W
Therefore, the minimum power delivered by the motor to the lift is 34500 watts, which corresponds to option 'C'.
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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)a)23000b)20000c)34500d)23500Correct answer is option 'C'. Can you explain this answer?
Question Description
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)23000b)20000c)34500d)23500Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 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)23000b)20000c)34500d)23500Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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)23000b)20000c)34500d)23500Correct answer is option 'C'. Can you explain this answer?.
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)23000b)20000c)34500d)23500Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 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)23000b)20000c)34500d)23500Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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)23000b)20000c)34500d)23500Correct answer is option 'C'. Can you explain this answer?.
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