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`        An escalator joins one floor with another one 8.20 m above. The escalator is 13.3 m long and moves along its length at 62.0 cm/s. (a) What power must its motor deliver if it is required to carry a maximum of 100 persons per minute, of average mass 75.0 kg? (b) An 83.5-kg man walks up the escalator in 9.50 s. How much work does the motor do on him? (c) If this man turned around at the middle and walked down the escalator so as to stay at the same level in space, would the motor do work on him? If so, what power does it deliver for this purpose? (d) Is there any (other?) way the man could walk along the escalator without consuming power from the motor?`
2 years ago

Arun
23769 Points
```							According to conservation of energy Work done to carry persons = change in potential energy Work = mgh₁ - mgh₂ Initially they were at ground level h₁ = 0 Hence, Work = -m(-g)h₂ = mgh₂ = 100 × 75× 9.8 × 8.2 = 602700J Power = Work/time = 602700/60 = 10045W = 10.045kW RegardsArun (askIITians forum expert)
```
2 years ago
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• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions

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