Measuring average power during a jump

[RoyalCat]

Homework Statement

At your disposal are a scale and a measuring tape. You are not allowed to jump on the scale.
How would you go about measuring the average power you exert on the floor?

Homework Equations

$$P_{average}=\frac{\Delta E_{tot}}{\Delta t}$$

The Attempt at a Solution

I am honestly lost. Without a way to measure the time it takes to execute the jump, how am I supposed to calculate the average power? I mean, calculating the total energy difference is fairly straight-forward. Measure the height of the person's center of mass at the top of the jump (Momentary rest) and say that all you've gained is potential height energy (A good choice for a plane of reference here would be the height of the center of mass in the knees-bent position), and bam, you're done.

But how do I use that change in energy to calculate the average power?

A hint at the right way to approach this would be terrific.

With thanks in advance, Anatoli.

 

[LowlyPion]

Well you can figure the impulse can't you from the height?

That gives you the I = F * T = Δmv

You can get the maximum force reading from the scale, ... maybe deduce the time from that?

 

[xavior13]

Can't you use delta y= Vi(t)+at^2/2? Vi=0, a=9.8, delta y measured solve for t then multiply t times two for the whole trip then divide delta E by delta t? seems reasonable although i usually overlook some detail -_-. tell me when you solve it. good luck.

 

[RoyalCat]

EDIT:
Maybe it's one of those 'think outside the box' exercises? I could measure the time by dropping the scales from a measured height, until the time it takes them to drop, matches the time it takes the jumper to straighten out, and from there, since I have the time it takes to transfer the energy, the rest of the calculations are trivial
?

xavior13, there is no way to measure the time it takes to execute the jump. And your acceleration during a jump is FAR from constant, and FAR from just free-fall.

LowlyPion, we're not allowed to jump on the scale. So it's only there to measure our mass (Or for some other creative use which I haven't figured out yet.)

Calculating $$\Delta mv$$ is tricky as well. I can't see how I can take care of the whole height energy conundrum. Would setting my plane of reference at the height of the center of mass when the jumper is on the tips of his toes do me any good? Would it be correct to claim, that at that point during his jump, all of his negative height energy has been converted into kinetic energy?
$$W=\Delta E_k$$
But that would still leave the problem of how long it took him to straighten out, though. >>

 

[LowlyPion]

LowlyPion, we're not allowed to jump on the scale.

I wasn't suggesting jumping on the scale.

I was suggesting to jump off the scale.

 

[RoyalCat]

I wasn't suggesting jumping on the scale.

I was suggesting to jump off the scale.

One of two things is eluding me here. It's either the irony, or the difference. Care to clarify?

 

[LowlyPion]

One of two things is eluding me here. It's either the irony, or the difference. Care to clarify?

If you stand on the scale and jump off observing the maximum deflection of the needle and determining the height that you jump, doesn't that give you a way to determine the impulse and the maximum force?

F * T = Δmv

 

[RoyalCat]

If you stand on the scale and jump off observing the maximum deflection of the needle and determining the height that you jump, doesn't that give you a way to determine the impulse and the maximum force?

F * T = Δmv

I'd say no. You would only be able to measure the maximal deflection (Unless you have a video camera that can help you draw a graph of $$F(t)$$ which you would then integrate), and that's only if the scales can get to that reading at all, or fast enough (Don't you end up exerting several thousand Newtons during a jump?).

The maximal deflection doesn't tell us much about the average force.
And I'm fairly certain they meant jumping on (Or off. (: ) the scale in any sort of way is not allowed.

 

[xavior13]

lol yeah i think Lowly is reading too much into technicalities.

try this: avg power= square root of 4.9 times mass times maximum height times 9.81

 

[davieddy]

Measure how far you crouch down before you jump.
Assume constant acceleration, and you can calculate the time taken
and acceleration during the actual jumping (straightening the legs).

 

[RoyalCat]

Measure how far you crouch down before you jump.
Assume constant acceleration, and you can calculate the time taken
and acceleration during the actual jumping (straightening the legs).

Interestingly enough, that's almost word-for-word what the full wording of the problem suggested I do (Minus the 'assume constant acceleration' bit), but I don't really understand how assuming constant acceleration let's me find the time taken.

All that does, is give me a complicated way to calculate the work done, and I think it's much better to just look at the crouched position compared to the top of the jump position to see what the work done was.

 

[xavior13]

im sry i made a mistake.

solve it with this: avg power=sq root of 4.9 times mass times sq root of max height times 9.81

its very simple. potential energy at top=work= mgh. Avg power=mgh/time, time= sq of 2h/a then simplify to wat i typed above and solve. a=freefall acceleration.

tell me wat u get, and do u have the answer to the problem?

 

[davieddy]

He crouches to a depth d, accelerates to speed v in time t and then ascends
under gravity to a height h.

mv^2/2=mgh
So we know v.

d=vt/2 so t=2d/v

Work done during jump = mv^2/2 + mgd

Power = work/t

 

[RoyalCat]

He crouches to a depth d, accelerates to speed v in time t and then ascends
under gravity to a height h.

mv^2/2=mgh
So we know v.

d=vt/2 so t=2d/v

Work done during jump = mv^2/2 + mgd

Power = work/t

Bleh, I thought that was the solution for a second, but all that tells us is the power of gravity, since we don't know how long it takes him to actually get from the initial state, the crouch at depth d, to the final state, the upright position with enough velocity to reach height h, we don't know how long it actually takes him to do the work.

I'm really starting to lean towards my idea of dropping the scale from such a height, so that the time it takes it to fall is equal to the time it takes the jumper to straighten out, and then we can find out the time during which he does work.

xavior, when you're jumping, you are NOT accelerating at free-fall acceleration. That solution is completely incorrect.

 

[xavior13]

lol ok ok sheesh!

 

[davieddy]

we don't know how long it actually takes him to do the work.

As I said t = 2d/v
You know v and d
What's the problem?

 

[RoyalCat]

As I said t = 2d/v
You know v and d
What's the problem?

I do not understand that equation. Where did it come from, what does it mean? At what point does the jumper cover twice the depth of his crouch at the speed at which he leaves his crouch?

The jumper at no point moves at a constant velocity, I really can't make heads or tails of your equation.

 

[davieddy]

d = (u+v)t/2 (area under v-t graph or average velocity * time)
u=0

PLEASE try harder:)

 

[RoyalCat]

d = (u+v)t/2 (area under v-t graph or average velocity * time)
u=0

PLEASE try harder:)

Oooh, I think I see it now. Since I know that the initial speed is 0, and I can tell the final speed from the height at the top of the jump, I can see what the average acceleration is, since I know the distance over which the jumper accelerates, and from that, I can see how long it took as well.

Your equation still makes very little sense to me, though. Thanks anyway. :3

 

[davieddy]

Oooh, I think I see it now. Since I know that the initial speed is 0, and I can tell the final speed from the height at the top of the jump, I can see what the average acceleration is, since I know the distance over which the jumper accelerates, and from that, I can see how long it took as well.

Your equation still makes very little sense to me, though. Thanks anyway. :3

If you read Dr Al's "sticky" post on formulae,
you will see "average velocity = displacement / time"
and for constant acceleration,
"average velocity = (u+v)/2"

It follows that d = (u+v)/2 * t.
I don't know why he didn't include this in his list of constant acceleration
equations.
Perhaps he overestimated the intelligence of the average punter here :)


David
(with attitude)

 

[RoyalCat]

If you read Dr Al's "sticky" post on formulae,
you will see "average velocity = displacement / time"
and for constant acceleration,
"average velocity = (u+v)/2"

It follows that d = (u+v)/2 * t.
I don't know why he didn't include this in his list of constant acceleration
equations.
Perhaps he overestimated the intelligence of the average punter here :)David
(with attitude)

Well, I can understand what the equation means, but how does it fit into the framework of the problem? Or rather, how do you see it fitting in? I can't quite follow your meaning.

 

[davieddy]

Well, I can understand what the equation means, but how does it fit into the framework of the problem? Or rather, how do you see it fitting in? I can't quite follow your meaning.

Power = work/time
Work = mg(d+h)
Time = 2d/v
mv^2/2 = mgh

You measure d and h with the tape measure, and m with the scales.
As usual in this forum, you assume g is known.

 

[RoyalCat]

Power = work/time
Work = mg(d+h)
Time = 2d/v
mv^2/2 = mgh

You measure d and h with the tape measure, and m with the scales.
As usual in this forum, you assume g is known.

And why is the time it takes him to rise equal to 2d/v? I'm sorry if I'm starting to sound like a broken record, but I honestly don't understand how you got to that conclusion.

The acceleration during the jump is not constant, so the average velocity is unknown, and you can't say it's just $$\frac {\sqrt{2gh}}{2}$$. At best, you'd know the velocity at the point of lift-off thanks to considerations of energy, no?

I've found my own way.
The work performed by the jumper is indeed $$mg(h+d)$$
But it's also equal to the integral of $$\Sigma \vec F(y)$$ during the jump, which is (If we set our plane of reference at the straight-legged position) from -d to 0.
And that, we can equate that to $$\bar Fd$$
$$V_0=\frac{\bar F}{m}*t$$
$$t=\frac{mV_0}{\bar F}$$

I'm really curious about your solution, though.

 

[davieddy]

The acceleration during the jump is not constant

Sorry can't read your black boxes.
As you have noted (about 10 times) we need to know the time he spends jumping
(in contact with the floor).

If the acceleration was varying, this time might be anything.
Constant acceleration is both reasonable and simple. (He exerts a constant
force during his jump, namely as hard as he can push on the floor with his legs).

 

[RoyalCat]

Sorry can't read your black boxes.
As you have noted (about 10 times) we need to know the time he spends jumping
(in contact with the floor).

If the acceleration was varying, this time might be anything.
Constant acceleration is both reasonable and simple. (He exerts a constant
force during his jump, namely as hard as he can push on the floor with his legs).

That is simply false. I'm sorry, but as convenient of a premise that may be to work with, it is in no way grounded in reality.

I appreciate the help, and apologize for any offense my tone may have been responsible for. :)

 

[davieddy]

That is simply false. I'm sorry, but as convenient of a premise that may be to work with, it is in no way grounded in reality.

I appreciate the help, and apologize for any offense my tone may have been responsible for. :)

What is false? Constant acceleration of his c of m during the jumping?

You are making the same assumption when you say work = Fd
and mv = (F-mg)t.

BTW although "F" is the force exerted by him on the floor (and vice versa)
neither are doing any work on each other since the floor isn't moving.

If it's any help (and I am beginning to doubt it):

If mv^2/2 = Fd
and mv = Ft

then dividing the first equation by the second, we find

v/2 = d/t

 

[RoyalCat]

What is false? Constant acceleration of his c of m during the jumping?

You are making the same assumption when you say work = Fd
and mv = (F-mg)t.

BTW although "F" is the force exerted by him on the floor (and vice versa)
neither are doing any work on each other since the floor isn't moving.

If it's any help (and I am beginning to doubt it):

If mv^2/2 = Fd
and mv = Ft

then dividing the first equation by the second, we find

v/2 = d/t

You must've missed that I wrote $$\bar F*d$$.
$$\bar F$$ = average force over the distance d.
The average force times the distance, equals the integral of $$\Sigma \vec F(y) * d$$
Since both graphs, by definition, enclose the same 'area' under them, the work done.

And I'm afraid you're misunderstood a bit. $$\bar F =\bar N-mg$$
The normal force from the floor is what's doing the work, not the force that the jumper is exerting on the floor, just so we're clear on that.

I think I should clarify another point here (Heh, reading your post now, shows me this clarification was, in the end, for me).
$$W_{net}=\Delta E_k=\vec F_{net}*s$$ Where $$s$$ is the distance over which the net force has acted. For a force dependent on the distance, an integral is required.
$$W_{non conservative}=\Delta U+\Delta E_k$$

So if we're dealing with $$\bar F =\bar N-mg$$
Then it follows that:
$$W_{\bar F}=\frac{1}{2}mv^{2}$$
$$W_{\bar F}=\bar F * d$$

$$\vec J = \bar F * t$$
$$m\vec v = \bar F * t$$

$$\frac {t}{d}=\frac{\vec v}{\frac{1}{2}v^{2}}$$
$$t=\frac {2d}{v}$$
Well, would you look at that. (:

I just wish you'd have put it that way, in terms of work and impulse before just spurting the final equation without an explanation, you really confused me. :\
Thanks a bunch for helping me reach the solution. :)

Oh, and just on a side note, the floor is moving, ever so slightly. A minuscule amount of work is performed on it, but what we're looking at is the work done on the jumper, and not the floor. The latter is irrelevant to us, since for all intents and purposes, the floor does not move.

 

[davieddy]

I just wish you'd have put it that way, in terms of work and impulse before just spurting the final equation without an explanation, you really confused me. :\
Thanks a bunch for helping me reach the solution. :)

The "final equation" doesn't need force, work, energy, momentum,
and doesn't even explicitly include acceleraration.

Have a good day.
Bye (before I get barred again for my attitude)

David

PS I thought of something else to say, but mercifully for both of us
I have temporarily forgotten it.
Watch this space.

 

[RoyalCat]

The "final equation" doesn't need force, work, energy, momentum,
and doesn't even explicitly include acceleraration.

Have a good day.
Bye (before I get barred again for my attitude)

David

PS I thought of something else to say, but mercifully for both of us
I have temporarily forgotten it.
Watch this space.

I don't think I like that implied threat, in light of my gratitude and apology.

Hrrmph, something is bothering me now, though. $$\bar F$$ with respect to time, is not necessarily the same as $$\bar F$$ with respect to the distance. Yuck.
I think my solution's borken again.

 

[RoyalCat]

Sorry for the double post, but I still can't wrap my head around this problem. Any help would be greatly appreciated. :)