Power required to climb a 20-m-tall building in 55s

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A 90 kg firefighter carrying 40 kg of gear needs to climb a 20-m-tall building in 55 seconds, resulting in a total mass of 130 kg. The potential energy at the top is calculated using the formula U = mgy, yielding 25,480 Joules. Dividing this energy by the time of 55 seconds gives a power requirement of 463 Watts. This calculation assumes perfect efficiency, though real-world conditions would result in energy loss to thermal energy. The discussion highlights the importance of considering efficiency when calculating power needs for such tasks.
ChetBarkley
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Summary:: A 90 kg firefighter needs to climb the stairs of a 20-m-tall building while carrying 40kg of gear. How much power does he need to reach the top of the building in 55s.

So first the total mass of our system is 130 kg. Using this mass, I found the potential energy the firefighter would have at the top of the building. Using U = mgy, I got 25480 J then knowing that power is just Jules over time I divided this by 55s and got 463 Watts. Is this correct?
 
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Looks good to me. That assumes perfect efficiency, but that's OK. In real life, much of the work done by the firefighter will go to thermal energy.
 
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Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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