Power needed to lift the average American man 3 stories

  • Thread starter Rewilding
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  • #1
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I'm not a maths person, so bear with me, but i'm trying to prove that with 3.5kW you could lift the average american male (89kg) to a height (take just as displacement) of 3 storeys (about 9.14 meters) in under 3 seconds.

Is somebody able to calculate this for me?

Thanks!
 

Answers and Replies

  • #2
Chandra Prayaga
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Well, the data is all there. Calculate the work done, which is mgh, and divide that by the time to get the power.
 
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  • #3
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I've got 2.28 seconds, so that all works - that's great thank you.

Is there no need to take into account acceleration?
 
  • #4
SteamKing
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I've got 2.28 seconds, so that all works - that's great thank you.

Is there no need to take into account acceleration?
Yeah, but what's the power? 2.28 sec. is a duration in time.
 
  • #5
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Is there no need to take into account acceleration?
Apparently initial velocity is zero, so yes, acceleration must be taken into account.
 
  • #6
OmCheeto
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Apparently initial velocity is zero, so yes, acceleration must be taken into account.
Off the top of my head, I disagree.
Can you explain why must acceleration be taken into account?
 
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  • #7
russ_watters
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Off the top of my head, I disagree.
Can you explain why must acceleration be taken into account?
It may matter with some methods, but may not matter with others. So without more detail, I like post #2 as a starting point.
 
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  • #8
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Yeah, but what's the power? 2.28 sec. is a duration in time.
Well, the power is the 3500W.
3500W x 2.28s = 7980J
89kg x 9.81m/s2 x 9.14m = 7980J
 
  • #9
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Yeah, but what's the power? 2.28 sec. is a duration in time.
3.5 kW as stated in the first post.
I've got 2.28 seconds, so that all works - that's great thank you.

Is there no need to take into account acceleration?
It works if you consider the average power and don't care about the precise acceleration profile. If 3.5 kW is the maximal power, you have to take acceleration into account. To make it worse, acceleration then depends on the velocity. I'll solve it here as it is probably beyond the scope of the original problem.
$$a = \frac{Pv}{m} - g$$
Where P is the power, v is velocity, m is the mass, g is the gravitational acceleration and a is the acceleration of the human.
That equation is solved by $$v=\frac{P}{mg} \left(1-e^{-tg^2m/P}\right)$$
Integrating gives
$$x(t)=\frac{Pt}{mg} - \frac{P^2}{m^2g^3} \left( 1 - e^{-tg^2m/P}\right)$$
The left term is 11.8 meters, and follows the previous calculation. The right term is 1.55 meters - it is a reduction of the height reached due to the initial low velocity.
The difference is still more than 10 meters, so yes, it is possible even if we take acceleration into account.
 
  • #10
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Off the top of my head, I disagree.
Can you explain why must acceleration be taken into account?
Doesn't the 2.28s take into account only the weight force moved through the 9.14m?
Wouldn't the required acceleration add to that, thus increasing the time?
(My final conclusion is that it is possible, assuming constant acceleration.)
 
  • #11
russ_watters
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3.5 kW as stated in the first post.
That's not what Steamking was asking. The method in post #2 was to calculate the average power required based on the provided height, weight and time, then compare it with the 3.5 kW limit. I see this as the simpler method.

[Edit] I also prefer it because given the [lack of] constraints, it provides a surer path to the correct answer. The provided power may be high enough that the constant acceleration assumption provides the right answer, but if the provided power was 2.8 kW it wouldnt.
 
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  • #12
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For constant acceleration:

Edit: How NOT to do it (see below):

9.14 = 0.5a32
a = 2.03m/s2
Fa = 89 x 2.03 = 181N
Fg = 89 x 9.81 = 873N
FTOT = 181 + 873 = 1050N
Work = 1050N x 9.14m = 9600J
Power = 9600J/3s = 3200W < 3.5kW
Q.E.D.
 
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  • #13
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@insightful: that is the average power, not the maximal one. If you just care about average power, there is no need to use constant acceleration, quasi-instantaneous acceleration is faster then.

A constant acceleration of 2.03 m/s^2 leads to a final velocity of 6.09 m/s, multiplied by the force of 1050 N you need a power of more than 6 kW at the end.
 
  • #14
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A constant acceleration of 2.03 m/s^2 leads to a final velocity of 6.09 m/s, multiplied by the force of 1050 N you need a power of more than 6 kW at the end.
Mea culpa, of course, you're right.
 
  • #15
Chandra Prayaga
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mgh/Δt gives the minimum power required. It is assumed there that the upward force is just balancing the weight throughout. If you need acceleration, then certainly, the force required in the upward direction is more, and that increases the work done, and therefore the power.

I am worried about the starting equation given by mfb:

a = (Pv/m) - g

The units don't match.
 
  • #16
russ_watters
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mgh/Δt gives the minimum power required. It is assumed there that the upward force is just balancing the weight throughout. If you need acceleration, then certainly, the force required in the upward direction is more, and that increases the work done, and therefore the power.
Actually, one reason I liked your method better is that it allows for the energy of the acceleration to be recovered when you decelerate! Which it does by itself if you let gravity decelerate him.
 
  • #17
jbriggs444
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You can reduce maximum power needed for the "acceleration required" version by allowing the man to be chopped up before being lifted, piecemeal.
 
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  • #18
OmCheeto
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You can reduce maximum power needed for the "acceleration required" version by allowing the man to be chopped up before being lifted, piecemeal.
Methinks that we may be scaring the bejeezits out of @Rewilding , with all of our smart talking.

And I haven't even interjected yet, that "mgh" is merely a non-spacefaring, earthling-centric approximation.

Oops! There we go again..... :redface:
 
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  • #19
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I am worried about the starting equation given by mfb:

a = (Pv/m) - g

The units don't match.
Maybe: a = P/(vm) - g (typo)?

The rest looks OK.
 
  • #20
russ_watters
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You can reduce maximum power needed for the "acceleration required" version by allowing the man to be chopped up before being lifted, piecemeal.
As an added benefit, that mitigates safety concerns regarding high initial acceleration, enabling a closer approximation of the constant power scenario.
 
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  • #21
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Using suvat and F=ma, and 3500W maximum power available, I get:

1. Accelerate at 4m/s2 up to 2.84m/s in 0.71s to 1.01m high.

2. Then accelerate at 1m/s2 up to 3.64m/s in an additional 0.79s to 3.57m high.

3. Then accelerate at 0.3m/s2 up to 3.89m/s in an additional 0.83s to 6.69m high.

4. Remain at 3.89m/s for the remaining 2.45m which takes 0.63s to the final 9.14m high.

5. Total time: 0.71 + 0.79 + 0.83 + 0.63 = 2.96s. (<3s)
 
  • #22
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mgh/Δt gives the minimum power required. It is assumed there that the upward force is just balancing the weight throughout. If you need acceleration, then certainly, the force required in the upward direction is more, and that increases the work done, and therefore the power.

I am worried about the starting equation given by mfb:

a = (Pv/m) - g

The units don't match.
Right, v should be in the denominator.
a = P/(mv) - g

Looks ugly, so I simulated it numerically. I get 11.2 meters, where I guess ~0.2 meters are uncertainty from the finite (but adaptive: smaller at the start) step size.
9.14 meters are reached after about 2.48 seconds.
 
  • #23
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Well this has been an amazing response! OmCheeto is right of course - it scares me what all your minds might be capable of! Only, of course, if you could all agree on a method...

This has been amazingly helpful anyway, whether we take acceleration into account or not it seems that under 3 seconds is a legitimate statement.

Thanks again all!
 
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  • #24
OmCheeto
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Well this has been an amazing response! OmCheeto is right of course - it scares me what all your minds might be capable of! Only, of course, if you could all agree on a method...
Being the most simple minded of the respondents, I chose the simplest method, which was described in Post #2.
This has been amazingly helpful anyway, whether we take acceleration into account or not it seems that under 3 seconds is a legitimate statement.

Thanks again all!
You're welcome.

And regarding my comment in post #18, if you should ever get into astrophysics, pe ≠ mgh.
This had me quite confused, this January, when I started following the Dawn space mission.
 
  • #25
sophiecentaur
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Methinks that we may be scaring the bejeezits out of @Rewilding , with all of our smart talking.
Yebbut, the OP does demand a proper answer and the 'simple' / obvious answer is not the right one. The KE at the top is non-zero if the journey is as fast as possible and the Power is limited. He's lifted the lid of a Pandora's Box - as so many OPs do. Lesson: do not tweak the tails of experts because they will turn round and confuse you.
 
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