Calculate Avg Power Output for 75kg Person Walking 100ft in 3min

[Madelin Pierce]

Homework Statement

Calculate average power output for a 75 kilogram person walking from ground level up stairs to a final height of 100 feet (then stops) in a time of 3 minutes

Homework Equations

P= work/ time or P=Fv

The Attempt at a Solution

I thought this was more of an energy problem where KE and PE are zero in the beginning, and then PE is m*g*h and KE is still zero. But I also think it's just finding average velocity and then finding acceleration and then power. But is distance the height though? I'm confused by all this.

 

[DaveE]

What is the kinetic and potential energy at the top compared to the bottom? How long did it take to do that?
Given the information provided how would you find velocity or acceleration?

 

[haruspex]

I thought this was more of an energy problem where KE and PE are zero in the beginning, and then PE is m*g*h and KE is still zero

I believe that is what is wanted. Any KE that had to be created to get going can be absorbed by "coasting" the last bit of the ascent, so there is no net expenditure on that.

 

[Madelin Pierce]

What is the kinetic and potential energy at the top compared to the bottom? How long did it take to do that?
Given the information provided how would you find velocity or acceleration?

Kinetic energy would be zero because the problem indicates the word stop as in rest. Potential energy is there as the person is higher above ground than before. I do think it's a P= energy/time problem

 

[haruspex]

Kinetic energy would be zero because the problem indicates the word stop as in rest. Potential energy is there as the person is higher above ground than before. I do think it's a P= energy/time problem

Yes.
DaveE may have been suggesting adding the KE required to go that distance in that time but a) we do not know the length of the path and b), as I indicated, the energy invested in it is recoverable.

 

[DaveE]

Yes.
DaveE may have been suggesting adding the KE required to go that distance in that time but a) we do not know the length of the path and b), as I indicated, the energy invested in it is recoverable.

No I was suggesting that if you compare zero KE at the beginning to zero KE at the end then you haven't stored any energy kinetically. Not so with potential energy.

 

[haruspex]

No I was suggesting that if you compare zero KE at the beginning to zero KE at the end then you haven't stored any energy kinetically. Not so with potential energy.

Ok, thanks for clarifying, but it is a bit subtler than you make out.
If you are taking initial KE as zero then some energy must be subsequently invested in KE to reach the required speed. If the speed is maintained all the way to the top then some KE is wasted, so there is a net cost of KE that should be included in the average power. To avoid this, it is necessary to model the process as coasting (i.e. shutting off power) as soon as the KE is enough to reach the top without further effort.

 

[DaveE]

Ok, thanks for clarifying, but it is a bit subtler than you make out.
If you are taking initial KE as zero then some energy must be subsequently invested in KE to reach the required speed. If the speed is maintained all the way to the top then some KE is wasted, so there is a net cost of KE that should be included in the average power. To avoid this, it is necessary to model the process as coasting (i.e. shutting off power) as soon as the KE is enough to reach the top without further effort.

Oh OK. I thought this was a homework problem. The ONLY information provided pertains to the initial and final states, I think it is safe to assume that everything in between can be ignored for this problem. However, you are absolutely correct, coasting uphill is a good way to stop.

 

[haruspex]

The ONLY information provided pertains to the initial and final states

Well, not quite. We know the height ascended and the time taken. This implies a minimum average speed, but since it is unlikely to be vertically upwards that would be an underestimate.

 

[DaveE]

I thought this was more of an energy problem where KE and PE are zero in the beginning, and then PE is m*g*h and KE is still zero. But I also think it's just finding average velocity and then finding acceleration and then power. But is distance the height though? I'm confused by all this.

You can solve this using either approach you described, they both result in the same formula. The average velocity you would use is just the vertical velocity, that is the only direction that work is done. Since there is no opposing force to the horizontal velocity component.
So P = (ΔPE)/Δt = (mgΔh)/Δt or P = F*v = (mg)*(Δh/Δt).