- #1

- 3

- 1

Is somebody able to calculate this for me?

Thanks!

- Thread starter Rewilding
- Start date

- #1

- 3

- 1

Is somebody able to calculate this for me?

Thanks!

- #2

Chandra Prayaga

Science Advisor

- 650

- 149

- #3

- 3

- 1

Is there no need to take into account acceleration?

- #4

SteamKing

Staff Emeritus

Science Advisor

Homework Helper

- 12,796

- 1,668

Yeah, but what's the power? 2.28 sec. is a duration in time.

Is there no need to take into account acceleration?

- #5

- 569

- 85

Apparently initial velocity is zero, so yes, acceleration must be taken into account.Is there no need to take into account acceleration?

- #6

OmCheeto

Gold Member

- 2,130

- 2,577

Off the top of my head, I disagree.Apparently initial velocity is zero, so yes, acceleration must be taken into account.

Can you explain why must acceleration be taken into account?

- #7

russ_watters

Mentor

- 19,875

- 6,295

It may matter with some methods, but may not matter with others. So without more detail, I like post #2 as a starting point.Off the top of my head, I disagree.

Can you explain why must acceleration be taken into account?

- #8

- 569

- 85

Well, the power is the 3500W.Yeah, but what's the power? 2.28 sec. is a duration in time.

3500W x 2.28s = 7980J

89kg x 9.81m/s

- #9

mfb

Mentor

- 34,822

- 11,000

3.5 kW as stated in the first post.Yeah, but what's the power? 2.28 sec. is a duration in time.

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.

Is there no need to take into account acceleration?

$$a = \frac{Pv}{m} - g$$

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

- 569

- 85

Doesn't the 2.28s take into account only the weight force moved through the 9.14m?Off the top of my head, I disagree.

Can you explain why must acceleration be taken into account?

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

Mentor

- 19,875

- 6,295

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.3.5 kW as stated in the first post.

[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:

- #12

- 569

- 85

For constant acceleration:

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

9.14 = 0.5a3^{2}

a = 2.03m/s^{2}

F_{a} = 89 x 2.03 = 181N

F_{g} = 89 x 9.81 = 873N

F_{TOT} = 181 + 873 = 1050N

Work = 1050N x 9.14m = 9600J

Power = 9600J/3s = 3200W < 3.5kW

Q.E.D.

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

9.14 = 0.5a3

a = 2.03m/s

F

F

F

Work = 1050N x 9.14m = 9600J

Power = 9600J/3s = 3200W < 3.5kW

Q.E.D.

Last edited:

- #13

mfb

Mentor

- 34,822

- 11,000

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

- 569

- 85

Mea culpa, of course, you're right.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

Chandra Prayaga

Science Advisor

- 650

- 149

I am worried about the starting equation given by mfb:

a = (Pv/m) - g

The units don't match.

- #16

russ_watters

Mentor

- 19,875

- 6,295

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.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.

- #17

jbriggs444

Science Advisor

Homework Helper

2019 Award

- 9,208

- 3,904

- #18

OmCheeto

Gold Member

- 2,130

- 2,577

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.....

- #19

- 569

- 85

Maybe: a = P/(vm) - g (typo)?I am worried about the starting equation given by mfb:

a = (Pv/m) - g

The units don't match.

The rest looks OK.

- #20

russ_watters

Mentor

- 19,875

- 6,295

As an added benefit, that mitigates safety concerns regarding high initial acceleration, enabling a closer approximation of the constant power scenario.

- #21

- 569

- 85

1. Accelerate at 4m/s

2. Then accelerate at 1m/s

3. Then accelerate at 0.3m/s

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

5.

- #22

mfb

Mentor

- 34,822

- 11,000

Right, v should be in the denominator.

I am worried about the starting equation given by mfb:

a = (Pv/m) - g

The units don't match.

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

- 3

- 1

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!

- #24

OmCheeto

Gold Member

- 2,130

- 2,577

Being the most simple minded of the respondents, I chose the simplest method, which was described in Post #2.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...

You're welcome.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!

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

Science Advisor

Gold Member

- 25,074

- 4,761

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.Methinks that we may be scaring the bejeezits out of @Rewilding , with all of our smart talking.

- Last Post

- Replies
- 9

- Views
- 1K

- Last Post

- Replies
- 9

- Views
- 9K

- Last Post

- Replies
- 25

- Views
- 5K

- Last Post

- Replies
- 1

- Views
- 4K

- Last Post

- Replies
- 4

- Views
- 4K

- Last Post

- Replies
- 1

- Views
- 4K

- Replies
- 6

- Views
- 3K

- Replies
- 3

- Views
- 8K

- Replies
- 7

- Views
- 2K

- Replies
- 3

- Views
- 2K