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A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to: (1 HP = 746 W, g = 10 ms-2)
  • a)
    1.5 ms-1
  • b)
    1.9 ms-1
  • c)
    1.7 ms-1
  • d)
    2.0 ms-1
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A 60 HP electric motor lifts an elevator having a maximum total load c...
Ftotal = Mg + friction
= 2000 × 10 + 4000
= 20,000 + 4000 = 24,000 N
P = F × v
60 × 746 = 24,000 × v
⇒ v = 1.86 m/s ≈ 1.9 m/s
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Community Answer
A 60 HP electric motor lifts an elevator having a maximum total load c...
Given Data
- Power of the motor: 60 HP
- Load capacity: 2000 kg
- Frictional force: 4000 N
- Acceleration due to gravity (g): 10 m/s²
- Conversion: 1 HP = 746 W
Step 1: Calculate the Power in Watts
- Total power of the motor in watts:
- 60 HP x 746 W/HP = 44760 W
Step 2: Calculate the Weight of the Elevator
- Total weight (W) of the elevator (including the load):
- W = mass x g
- W = 2000 kg x 10 m/s² = 20000 N
Step 3: Determine the Net Force Acting on the Elevator
- Net force (F_net) is calculated as:
- F_net = Power / Speed
- However, the total force also includes the frictional force:
- F_net = Weight + Friction
- F_net = 20000 N + 4000 N = 24000 N
Step 4: Calculate the Speed of the Elevator
- We know that Power = Force x Speed. Rearranging gives:
- Speed = Power / Force
- Here, Force = F_net = 24000 N
- Speed = 44760 W / 24000 N = 1.865 m/s, which rounds to approximately 1.9 m/s.
Conclusion
The speed of the elevator at full load, considering the friction, is close to 1.9 m/s, confirming that option 'B' is the correct answer.
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A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to: (1 HP = 746 W, g = 10 ms-2)a)1.5 ms-1b)1.9 ms-1c)1.7 ms-1d)2.0 ms-1Correct answer is option 'B'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to: (1 HP = 746 W, g = 10 ms-2)a)1.5 ms-1b)1.9 ms-1c)1.7 ms-1d)2.0 ms-1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to: (1 HP = 746 W, g = 10 ms-2)a)1.5 ms-1b)1.9 ms-1c)1.7 ms-1d)2.0 ms-1Correct answer is option 'B'. Can you explain this answer?.
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