# Simple work and energy problem [closed]

I have the following problem:

A man that weights 50kg goes up running the stairs of a tower in Chicago that is 443m tall. What is the power measured in watts if he arrives at the top of the tower in 15min?

So far I used this equation:

W = ΔEc + ΔEp + ΔEq

Where ΔEc is the variation of the kinetic energy, ΔEp is the variation of the potential energy (based on the height) and ΔEq is the energy based on friction.

There isn't an initial W and a ΔEq, so I have this equation:

ΔEc = - ΔEq

Am I doing something wrong? I am stuck here.

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## closed as off-topic by ACuriousMind, Kyle Kanos, BMS, David Z♦Sep 11 '14 at 18:22

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Power is energy per second. I would work out the energy change involved in ascending 443m and divide it by the time taken for the ascent. I don't think kinetic energy and losses due to friction are relevant. – John Rennie Sep 13 '12 at 14:42
So I only use potential energy? I thought there was kinetic energy as the man who is running in the tower has a velocity, or am I wrong? – Pacha Sep 13 '12 at 14:46
The question isn't worded particularly well. The change in potential energy is easy to calculate. The time over which the energy changed is given. The average power associated with this change in potential energy is just the change in potential energy divided by the time during which the change took place. – Alfred Centauri Sep 13 '12 at 15:00
Hint to jugde whether ΔEc is important: compare with ΔEp. It will be much much smaller. – Alexander Sep 13 '12 at 15:33
This can't be meaningfully answered unless you first declare what you're taking to be the system. If it's just the man, then you can't speak of potential energy because a single entity cannot possess potential energy. You must then speak of work done on the system by some agent (Earth, in this case). Either way, friction is irrelevant here because there's no displacement of the feet at the point of application of the friction force. The stairs do no work on the man. – user11266 Dec 12 '12 at 22:45

First, we have to get the weight of the person from his mass. 50 kg here on earth has a weight of 490 N. That force times the distance over which it is applied gives the total energy: 490 N x 443 m = 217 kJ. That amount of work was done in 15 minutes, which is 900 seconds. 217 kJ / 900 s = 241 W average over the 15 minutes.

Yes, it really is that simple.

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Assuming man stops on reaching at top, all the energy is now converted into potential energy. I am ignoring friction loss etc as no such information is available. Given this

Potential Energy = mgh

and Power = Energy/Time

Thus assuming g =9.8m/sec^2

Power = 241.18 Watt

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Assume he starts off stationary, and ends up stationary (he'll accelerate and decelerate in between, but that's by the by). So now you know ${\Delta}Ec$ .

Now, you say there is no initial $W$, but you don't say what $W$ is. I'm guessing it's work done. (and remember the relationship between work done, power and time). Initial work is meaningless: ${\Delta}Ec + {\Delta}Ep + {\Delta}Eq$ equals total work done, i.e. $W$.

And finally, where does the energy consumed in friction end up? Will that be significant relative to ${\Delta}Ep$?

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