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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)20000
- b)34500
- c)23500
- d)23000
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
An electric lift with a maximum load of 2000 kg (lift + passengers) is...
Constant velocity ⇒ a = 0
⇒ T = W + f
= 20000 + 3000
= 23000 N
⇒ T = W + f
= 20000 + 3000
= 23000 N
⇒ Power = Tv
= 23000 × 1 .5
= 34500 watts
= 23000 × 1 .5
= 34500 watts
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Community Answer
An electric lift with a maximum load of 2000 kg (lift + passengers) is...
To solve this problem, we need to use the concept of force and work.
First, let's calculate the force required to lift the maximum load of 2000 kg. The force can be calculated using the formula:
Force = mass * acceleration
The acceleration can be calculated using the formula:
acceleration = change in velocity / time
In this case, the change in velocity is 0 m/s (since the lift is moving at a constant speed) and the time is not given. Therefore, we can assume that the time is large enough for the lift to reach a constant speed.
So, the force required to lift the maximum load is:
Force = 2000 kg * acceleration
Next, let's calculate the work done by the electric lift. The work done can be calculated using the formula:
Work = force * distance
In this case, the distance is not given. Therefore, we need more information to calculate the work done.
Please provide the distance the electric lift travels.
First, let's calculate the force required to lift the maximum load of 2000 kg. The force can be calculated using the formula:
Force = mass * acceleration
The acceleration can be calculated using the formula:
acceleration = change in velocity / time
In this case, the change in velocity is 0 m/s (since the lift is moving at a constant speed) and the time is not given. Therefore, we can assume that the time is large enough for the lift to reach a constant speed.
So, the force required to lift the maximum load is:
Force = 2000 kg * acceleration
Next, let's calculate the work done by the electric lift. The work done can be calculated using the formula:
Work = force * distance
In this case, the distance is not given. Therefore, we need more information to calculate the work done.
Please provide the distance the electric lift travels.
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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)20000b)34500c)23500d)23000Correct answer is option 'B'. 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)20000b)34500c)23500d)23000Correct answer is option 'B'. 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)20000b)34500c)23500d)23000Correct answer is option 'B'. 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)20000b)34500c)23500d)23000Correct answer is option 'B'. 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)20000b)34500c)23500d)23000Correct answer is option 'B'. 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)20000b)34500c)23500d)23000Correct answer is option 'B'. 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)20000b)34500c)23500d)23000Correct answer is option 'B'. Can you explain this answer?.
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