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An electric lift with a maximum load of 2000kg (lift + passengers) is moving up with a constant speed of 1.5ms−1. The frictional force opposing the motion is 3000N. The minimum power delivered by the motor to the lift in watts is : (g=10ms−2)
- a)23000
- b)20000
- c)34500
- d)23500
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
An electric lift with a maximum load of 2000kg (lift + passengers) is...
Given:
Maximum load of lift, W = 2000 kg
Speed of lift, v = 1.5 m/s
Frictional force opposing motion, f = 3000 N
Acceleration due to gravity, g = 10 m/s^2
To find:
Minimum power delivered by motor to the lift in watts.
Formula used:
Power = force x velocity
Calculation:
1. First, we need to find the total force acting on the lift.
Total force, F = W x g + f
= 2000 x 10 + 3000
= 23000 N
2. Now, we can calculate the minimum power delivered by the motor to the lift using the formula mentioned above.
Power = F x v
= 23000 x 1.5
= 34500 watts
Therefore, the correct option is (c) 34500.
Explanation:
The lift is moving up with a constant speed of 1.5 m/s, which means the net force acting on it is zero. However, there is a frictional force opposing the motion, which means the motor has to deliver a certain amount of power to overcome this force and keep the lift moving at a constant speed. The total force acting on the lift is calculated by adding the weight of the lift and its passengers (W x g) to the frictional force opposing the motion (f). Finally, we use the formula for power to calculate the minimum power delivered by the motor to the lift.
Maximum load of lift, W = 2000 kg
Speed of lift, v = 1.5 m/s
Frictional force opposing motion, f = 3000 N
Acceleration due to gravity, g = 10 m/s^2
To find:
Minimum power delivered by motor to the lift in watts.
Formula used:
Power = force x velocity
Calculation:
1. First, we need to find the total force acting on the lift.
Total force, F = W x g + f
= 2000 x 10 + 3000
= 23000 N
2. Now, we can calculate the minimum power delivered by the motor to the lift using the formula mentioned above.
Power = F x v
= 23000 x 1.5
= 34500 watts
Therefore, the correct option is (c) 34500.
Explanation:
The lift is moving up with a constant speed of 1.5 m/s, which means the net force acting on it is zero. However, there is a frictional force opposing the motion, which means the motor has to deliver a certain amount of power to overcome this force and keep the lift moving at a constant speed. The total force acting on the lift is calculated by adding the weight of the lift and its passengers (W x g) to the frictional force opposing the motion (f). Finally, we use the formula for power to calculate the minimum power delivered by the motor to the lift.
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Community Answer
An electric lift with a maximum load of 2000kg (lift + passengers) is...
Fup =2000g + 3000
= 23000N
Minimum power = 
= 34500W
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An electric lift with a maximum load of 2000kg (lift + passengers) is moving up with a constant speed of 1.5ms−1. The frictional force opposing the motion is 3000N. The minimum power delivered by the motor to the lift in watts is : (g=10ms−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 2000kg (lift + passengers) is moving up with a constant speed of 1.5ms−1. The frictional force opposing the motion is 3000N. The minimum power delivered by the motor to the lift in watts is : (g=10ms−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 2000kg (lift + passengers) is moving up with a constant speed of 1.5ms−1. The frictional force opposing the motion is 3000N. The minimum power delivered by the motor to the lift in watts is : (g=10ms−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 2000kg (lift + passengers) is moving up with a constant speed of 1.5ms−1. The frictional force opposing the motion is 3000N. The minimum power delivered by the motor to the lift in watts is : (g=10ms−2)a)23000b)20000c)34500d)23500Correct answer is option 'C'. Can you explain this answer?.
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