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Power needed to lift the average American man 3 stories

  1. Nov 19, 2015 #1
    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!
     
  2. jcsd
  3. Nov 19, 2015 #2
    Well, the data is all there. Calculate the work done, which is mgh, and divide that by the time to get the power.
     
  4. Nov 19, 2015 #3
    I've got 2.28 seconds, so that all works - that's great thank you.

    Is there no need to take into account acceleration?
     
  5. Nov 19, 2015 #4

    SteamKing

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    Yeah, but what's the power? 2.28 sec. is a duration in time.
     
  6. Nov 19, 2015 #5
    Apparently initial velocity is zero, so yes, acceleration must be taken into account.
     
  7. Nov 19, 2015 #6

    OmCheeto

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    Off the top of my head, I disagree.
    Can you explain why must acceleration be taken into account?
     
  8. Nov 19, 2015 #7

    russ_watters

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    It may matter with some methods, but may not matter with others. So without more detail, I like post #2 as a starting point.
     
  9. Nov 19, 2015 #8
    Well, the power is the 3500W.
    3500W x 2.28s = 7980J
    89kg x 9.81m/s2 x 9.14m = 7980J
     
  10. Nov 19, 2015 #9

    mfb

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    3.5 kW as stated in the first post.
    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.
     
  11. Nov 19, 2015 #10
    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.)
     
  12. Nov 19, 2015 #11

    russ_watters

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    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.
     
    Last edited: Nov 19, 2015
  13. Nov 19, 2015 #12
    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.
     
    Last edited: Nov 19, 2015
  14. Nov 19, 2015 #13

    mfb

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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.
     
  15. Nov 19, 2015 #14
    Mea culpa, of course, you're right.
     
  16. Nov 19, 2015 #15
    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.
     
  17. Nov 19, 2015 #16

    russ_watters

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    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.
     
  18. Nov 19, 2015 #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.
     
  19. Nov 19, 2015 #18

    OmCheeto

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    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:
     
  20. Nov 19, 2015 #19
    Maybe: a = P/(vm) - g (typo)?

    The rest looks OK.
     
  21. Nov 19, 2015 #20

    russ_watters

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    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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