EZP

In raising 62kg by 0.8m the kinetic energy when  upright  at the point of lift off must match the attained potential energy at top of jump is  mass*height, ie

0.8*61 = 48.8J

Since energy is force times distance then

48.8  = f*0.2

hence f = 244nm(-2)


I have does this and the other question in rapier speed so there may be erros I hope not.

Regards

GfW

This problem can be done in terms of distances and accelerations, but it is much easier to do from an energy point of view.
By this method the increase in energy is equal to the work done. Here kinetic energy is zero before the jump and at the peak. Thus we need consider only potential energy. Let the initial PE be zero. in the following MKS units are used. m is mass of person, h is the total height of jump, x is the distance from initial position until he leaves the ground, g = 9.8 acceleration due to gravity, F is the desired force. Assumptions - force is constant. distances are measured to the center of gravity, and the configuration of the body does not change significantly.
-
PE = mgh = Work
Work = Fx
 
Therefore F = mgh/x = (61 9.8 1)/0.2 = 2989 or 3000 N

got my units wrong

f = 244 Newtons

During lift off, the force applied by the ground to the person causes the center of mass of the person to rise.

h = height of person

Treat person as a uniform distribution of mass (inhumane as that is!)

cm = center of mass height

cr = crouch distance

cm for full crouch at start of jump = (h - cr)/2

cm at no crouch as person leaves ground is h/2

Mass of person is moved through distance that cm changes = h/2 - (h - cr)/2 = cr/2

So work done by force of jump = F * cr/2

Max height that center of mass ascends after leaving ground is 0.8m

F * .1 = m*g*(0.8)

So F = 61 kg * 9.8 m/s/s * 0.8 / 0.1 = 4782 N

a yes I did make a mistake the potential energy is mgh I forgot good old gravity the potential energy is

0.8*61*9.8 = 478.24J

so the  force is 478.24/0.2 = 2391.2 N

 
Aburr: I am not right in saying that the the distance is 0.8 and not 0.8 + 0.2 = 1 ? I could be wrong I am rushing here.

aburr: I have slowed down here a bit and yes I agree with your analysis it is 1.0 not 0.8 .

Following aburr and GwynforWeb I see I neglected the PE gain from the bottom of the crouch too:

F*crouch/2 = m * g * (0.8 + crouch/2)

So F = 5380 N

My take is that "crouch" = distance head is lowered by in preparation for jump.

wytcom: I was thinking about the crouch/2 idea as well but it depends on how the jumper crouches I am not sure this problem is exactly defined.

The problem we have here is a problem for fair amount of points and all of us trying to do as fast as we can

OK GwynforWeb.  But the crouch issue is interesting since the ground actually does no work at all on the jumper.  After all the force applied by the ground on the jumper is applied at the same position throughout the jump.  This is not the situation that applies to work = force * distance.  We can cheat and use the work concept only by applying it here to the movment of the center of mass.

A simple model for this situation is a spring with mass.  If it is compressed against the ground then released it will jump up off the ground.  The problem can be solved without reference to work done by the ground since we can use the potential energy stored in the spring.  But we must track the position of the center of mass from the point of release to the heighest point of the jump to make this work.

I know what you are saying but where does the center of mass start , this is where I think there is a minor ambiguity.

Since I regard this as critical to the problem, I adopt the simplest model that includes it.

In this case, the model is that the person has uniform distribution of mass at all times.   So at the bottom of a crouch of length cr the center of mass is at height = (h - cr)/2.   h = height of person.

Then when the person is elongated at lift off (but still in touch with ground) the height of the center of mass is h/2.

Thus the "force" applied by the ground affects the movement of the center of mass over a distance of h/2 - (h - cr)2 = cr/2.

It looks to me like aburr is taking the specified crouch of 0.2 m to be the distance the center of mass moves through during the execution of the jump.  But, to me, that is not what is meant when the crouch for a jump is mentioned.  Plus if that is what is meant then the quality of the original question is significantly diminished.  For then the problem is the trivial one of pushing a mass through a specified distance to see how much energy is acquired.  The interest here is the subtlety of the concept of work applied to a jump and the way we can "cheat" to make it useful where it really is being misused.

I do also understand the spring idea which would drop it by 0.1 m but I can visualise crouching that would give different results. We have assume some thing and the spring is  a good idea, I think the problem needs a more precise defintion of the term crouching.

By treating the full crouch as the distance that the force from the ground is applied through we have an implicit model of the jumper as having massless legs!

You can see this by considering a person launched from a massless platform that applies a constant force through a fixed distance (the crouch) while the person stands erect.  This model yields the aburr and GwynnForWeb results.

The center of gravity of human being is around the chest area, it is suprisingly high up in the body. The point I am trying to make is that the problem is not exactly defined at the moment, I can see a crouch position albeit a little awkward where the center of gravity does pass through most of the 0.2m, I am in agreement with you though that the force cannot be considered to act over the full 0.2 m. I think we are basically in agreeement here probably the assumption of 0.1m is the best we can do. Remember we all did this question at brreakneck speed.

GwynforWeb,

This issue could be important.  Your answer and aburr's are both less than 3000 N while mine is more than 5000 N.  Suppose the jumper is on a brittle surface such as the ice of a lake and we are trying to calculate whether the jump up to 0.8 m could be made without applying a force greater than 4000 N to the ice.  The massless legs model would not be conservative enough to account for the real extra force that results from the reduced movement of the center of mass during the jump.  The uniform distribution of mass model may give too large of a value if as you say the legs aren't that significant as part of the body mass.  

But, if we are going to try to get real at all we will also have to abandon the notion that the force is constant.  This will also add to the peak force applied.

I agree 0.2m is to large a value as to what it is I do not know.

GwynforWeb,
Impressive.  I'd say you deserve the points on this one! I'm sticking to the easy theoretical part.

Yes, now I have a sore back and my kids think I am nuts!  :o)

wytcom: 'I guess we are having "phun"!   ", you are right! :o)

The essense of a physics problem is that appropriate assumptions are made at the start. If no assumptions are made every problem is impossible. Seth however has gone a bit far. If the mass is assumed to be all concentrated in the feet the answer is that an infinite force is required. Or to put it another way the person will not lift off the ground. As the person uncoils no energy is given to the body because all the mass is concentrated in the feet which remain firmly on the ground. Because the body now has no KE there is no possibility of the PE increasing.
It is not cheating to use energy considerations to solve this and similar problems. Just because it is simpler does not mean it is not legitemate.
The assumptions made are of course critical and should be stated. Of course the more realistic the assumptions the more realistic the answer. (and usually the more work required. See wytcom's solution where he relaxed one assumption).
They do tell the story of the physicist who was asked about whether of not it was really possible for a cow to jump over the moon. He replied, "Let us do a calculation. First, assume a spherical cow." ...

This is not true aburr.

In the *limit* that the center of mass goes to the ground, there is no change in the work done through that distance. As long as there is any non-zero mass above the feet, the work done is F*x and is mass-independent.

In a real-life situation, all of our answers are likely to be way off. There are many factors involved in such problems that we are not taking into consideration here, such as the fact that people do not exert equal forces at all times in a jump. They also don't all crouch in the same way, or have the same mass distribution. Like most "standard" physics problems, this problem is designed to have a definite answer, but it likely fails to do what it intended, since the only definite answer one can give is somewhat contrived.

My guess is that the problem should have said, "Assume the person's center of mass is lowered by 0.20 m while crouching"

This would have remove the ambiguity, but as it stands, I believe only the answer I gave can be proven correct.

-Seth

OK, aburr.  I was worried about applying Work = F * x here since it is derived for a particle moving though x while influenced by F.  But I see now that it is ok if we have a system of particles (person) and if x is the distance moved by the center of mass of the system while under the influence of the external force F.  

But I still say that the evaluation of the work should use a change in center of mass position = crouch/2 rather than the full crouch you used.

I just realized that I forgot to add the gravitational force, mg to my answer.

That would change it to about 3000 N instead.

But I think there is a problem with the energy conservation argument, so I will solve it using momentum conservation as well.

From the momentum perspective, his body has a momentum of mv where v is his velocity as he lifts off = sqrt(2gh).

Thus, he must acquire a momentum of (61)*sqrt(2*9.8*.80) = 240 kg*m/s over a period of time t.

During that time, the ground feels a force of mg + p/t, which is minimized when t is maximized. However, the maximum time to apply the force is constrained by the space through which he can move. Since this situation is reversible, we can look at what happens when he lands. What would be the maximum time he can take to stop his momentum as he lands on the ground, given that he ends up crouched by 0.2 m?

Now if his center of mass shifts by x, then the time to stop is 2x/v and the force becomes mg + mv^2/2x = mg + mgh/x.

Thus, the stopping time is maximized when x is maximal. Since it would not make sense for the center of mass to shift by more than the amount he crouches, the force is minimized by using 0.2 for x.

Therefore, we get mg(1+h/x) = 5mg = 5(61)(9.8) = 3000 N

which is the same as my other (corrected) answer, so I believe this is it, and isn't so unphysical after all.

-Seth

1) Conservation of momentum applies here only if you include the whole earth as part of the system.  Then momentum of earth + momentum of jumper = constant.

2) The work done by the ground on the jumper is 0.  This has to be so because the force applied by the ground to the jumper is always at the same position.

Any solution using the above 2 principles to solve this problem is suspect.  (Including mine, but I suspect mine is correct :-)

I did take the earth into account, though. The question asks for the force between the two, where F = dp/dt is the change in momentum with respect to time of each.

-Seth

There is actually one thing to realize if you want to consider a real situation. When a person performs a standing jump like that, they will move their arms to help propel themselves, effectively varying their mass distribution. Lowering the arms in mid air would raise the feet but not the center of mass, so I think the problem needs to be more clearly specified as to what is actually being measured.

You apparently don't understand the concept of a limit, aburr.

In the limit (from above) that the cm goes to ground level, the cm is *always* above the ground. I never stated at any time that the cm could be at the ground level. My analysis was performed in that limit, and was thus done over a continuous range that did not include the ground level itself, merely approaching it from above.

It's true that in that limit, the mass that could be lifted by the legs goes to zero. However, as long as it is greater than zero, the same amount of work is done by applying the same force though the same distance, irrespective of that mass. This means that that work done approaches a constant in that limit. Of course, it would also mean that the initial velocity of the upper body would go to infinity in that limit, but it never reaches that value. The point of what I'm doing is to analyze limiting cases, something that is very much done in real life to study boundary conditions and for verification testing.

If the problem had said "estimate" the force instead of what force "must" be exerted, it would be a different question entirely, and we would have use approximation techniques like what you are suggesting. However, that is not what the problem has asked for. When there is no other source of information to refer to, we must go by the precise wording of the problem. If it doesn't say what it means, then it is a badly worded problem.

-Seth

Also, remember what I said about the "real life" situation where a person would use his arms to gain additional advantage. It's much more difficult to do a standing jump when you can't move your arms. This is why teaching problems are designed to be unambiguous. The students should not be making any assumptions when solving the problems. It is not good practice for beginning students to do that. The best problems do not require any such assumptions, but still get the point across. The problem is that too much hand waving can make the explanations difficult to swallow, so there is no need to use problems that have built-in ambiguities, except for the more advanced students. And in that case it will be clear, because they will use such words as "estimate" or "approximate" in the problems.

Now, for example, in your analysis you *assumed* that the force is constant, when you should realize that that can be assumed without loss of generality in this problem based on the specific wording. The word "must" is what allows for this, and it can be proven. If you did not explain that to a student, how could you expect him to know he should assume that for such a problem. It's not even a reasonable assumption in real life, and very far from accurate. But for this problem, it can be *proven* to be a valid assumption for purposes of obtaining the correct answer. This is why we say assume "without loss of generality" in such a case.

-Seth

aburr, you say "...it is pemissable, even desirable, to simplify the problem as much as possible."   I accept the gist of this whole heartedly.  After all we are trying to demonstrate principles of physics using appropriate models.  

But, we need to take special care in selecting the "simplest" model.  I would revise this to say that the model needs to be the simplest one possible that includes the important basic ideas under study.  In the case of the jumper, we have internal energy being converted to kinetic energy  (like a massive spring compressed against the ground then released).  If our model of the jumper is equivalent to a particle of mass m being acted on by a force F through a distance x, then we have lost the special character of this situation.

That is why I point out that the ground does no work on the person jumping.  Yet we use an expression for work to solve the problem.  This hides the basic physics of the jump.  If we are teaching about work and use this problem to assist, then we mislead the student regarding the work concept.

To understand the jumper problem better, solve the problem of a massive spring compressed by c then released and observed to rise to a distance j from the ground to the bottom of the spring.  Take the mass of the spring as given, then find the force the spring exerts on the ground during the jump.  

As far as the assumptions go, I'm quite sure now that as long as the center of mass height does not change by more than 0.2 m while the person is on the ground, there is no difference in the calculation. The center of mass does not need to move the full 0.2 m in order to make the jump using only 3000 N of force on the ground. This is why I used the limit I did initially, to find the mass-distribution independent solution.

I realize some people may think that doing such a thing is crazy, but it is actually a sound technique for finding bounds. To demonstrate that the limit is indeed valid, consider a more realistic example. We would all agree that some part of the body must be lowered by 0.2 m in order to satisfy the crouch condition, let's say it's the head. Regardless of *any* other movements, it is possible to apply a total force of 2400 N to the head for the entire time it takes it to reach its orginal position, which converts enough potential energy to kinetic for the entire body.

Does the head get all that kinetic energy and transfer it to the body? Of course not, it would probably fly off if it did. The force is being applied everywhere throughout the body. But we don't need to consider the mass of each of those parts in order to determine the transfer of energy though, as I have shown. The extra 600 N on the ground comes from the weight of the person.

Now for a simplified version using a spring (a la wytcom):

Imagine that two masses, both weighing 300 N are stacked with springs beneath each. Thus, we have a mass on a spring on a mass on a spring on the ground. Assume the upper spring takes exactly 2700 N to compress, and the lower one takes 3000 N. Hence, neither spring begins compressed, since the upper one feels 300 N down and the lower one feels 600 N. Now, pressing down on the upper mass with a force of 2400 N will just begin to compress both springs simultaneously. Pressing down 0.2 m will do exactly 480 N*m of work, however the springs have also stored the lost potential energy of the masses. Thus, there is enough energy to get the masses back to their original positions, plus 480 joules extra to lift the enire contraption by 0.8 m.

Please try to convince yourself that the distribution of mass would not matter here, and apply it to the jumper's case. This is why my original analysis was done in the limit, to consider *all* cases, not one example. However, for some of you this example should help illustrate where your reasoning is going awry.

-Seth

seth, you say "Assume the upper spring takes exactly 2700 N to compress, and the lower one takes 3000 N. "

This does not seem to be consitent with the basic principle of a spring.  A spring is a device that compresses by a distance that is proportional to the force applied (F = k*x).  Saying it takes 2700N to compress a spring is inconsistent with this.  Any force will compress the spring some distance.  Double the force and it will compress it twice as much.  There is no minimum force to initiate compression of a spring.

I wasn't sure if I'd need to explain this, but you can certainly build a device like this. Also, any spring can be made to approximate this by compressing it ahead of time and preventing it from uncompressing beyond that. Regardless, it does not alter the argument I am making.

If you insist, we can use a pressurized hydraulic spring where the piston displaces as small a fraction of the total gas as we choose. That way, the change in force while displacing can be made as small as we wish. It is not an unphysical situation.

-Seth

A device similar to this is used in hydraulic lift chairs. You'll notice that they can lift and be pressed down with roughly the same force though its entire range of motion.

Sorry, I meant to say pneumatic, not hydraulic (gas, not liquid). But I'm sure you get the idea...

Of course the original question is open to interpretation. It would take a book to remove all possible questions. However to address the center of mass (cm) questions let us try this procedure. Let us define the jump distance as j, the distance from the original position that the head (or feet or cm) rises. In the original problem 0.8 m. Let c be the distance that the head is lowered in preparing for the jump, in the original problem 0.2 m, the crouch. (I took this to be the distance the cm was lowered.) However some objected that this view was open to question. Thus to increase generality let us define d as the distance that the cm was lowered. Add the parameter, r, defined as the ratio (<1) between the height of the head and the height of the cm at the start of the problem (before lowering for the jump). With reasonable assumptions about the body and its ability to fold we have
d = r c
I have done nothing to the original problem except add a parameter r that can be set to anything that the solver wishes. In my solution I assumed r = 1, wytcom assumed 0.2, SethHoyt postulated a value which approached 0 from above.
The basic physical principle involved is that of the conservation of energy. In this case the work done on the body is equal to the increase in potential energy. Here work is the force times the distance through which it is applied and the increase in potential energy is mgh. Thus F d = m g (d + j) or
-
F r c = m g (rc + j)
or

F = m g (1 + [j/r c])

Taking m as the original 61 kg and g = 9.8 we get
For my solution with r = 1  F = 2989 N which rounded to the two significant figures suggested by the number for the mass gives 3000 N.

If r = 0.5 as proposed by wytcom  F = 5380 N or 5400 N

It is also clear that if r approaches 0 as postulated by SethHoyt the force required increases without bounds, ie approaches infinity, by any standard way of finding the limit.
It is easy to see why F varies as shown with changing r. For the body to raise 0.8 m, a certain amount of work had to be done. The lower the cm the less distance through which the force has to operate, hence the greater the force necessary to do the required work.

aburr: "wytcom assumed 0.2"

Actually 0.5

aburr,

Good analysis.  

Nevertheless, the main issue outstanding is that the work done by the ground on the jumper is actually 0.  Your solution (and mine!) uses an invalid calculation of work somehow expecting it to apply here.  I actually think that the invalid calculation using the distance moved by the center of mass does give the correct answer (my answer).  But none of us has demonstrated why this should work.

For me, the significance here is that we are abusing the basic physics just to get an answer.  This is worse if we do it in the context of teaching students how to use the work concept.

aburr: "Here work is the force times the distance through which it is applied and the increase in potential energy is mgh. Thus F d = m g (d + j) ..."

As I have been trying to point out, this analysis is incorrect. If the body is not rigid, you cannot determine its change in kinetic energy based only on its rate of change in momentum (F=dp/dt) and cm displacement alone. For one thing, in a non-rigid body, some parts can move faster than others, obfuscating the total energy from any pure cm-based analysis.

If, for some reason, my spring example above is not sufficient to demonstrate this, I'll give another example to make the point even clearer why this analysis fails.

Imagine a person that can lift many times his weight so we can effectively ignore his own. If he lifts a mass m a distance d by applying a constant force F, he will increase its total energy by the work he does, F*d. Its potential energy is increased by mgd and its kinetic by (F-mg)*d. The F-mg is the extra it takes to lift vs. hold the mass steady. Neglecting his weight, the ground will feel a normal force of exactly F while he is lifting. This must be the case to cancel the lifting force F, otherwise he would accelerate due to the imbalance of the two forces.

If we attempt to determine the work done using just the force on the ground applied though the cm's displacement, we will get the correct answer, F*d. However, consider now that the person stands on a mass M of negligible thickness while lifting. This would not affect the total work done on the system. If the mass m was originally at his feet, the center of mass of the system rises by d*m/(m+M) during the lift. The ground now feels a normal force of F+Mg due to the additional weight of the mass M. The product of these is no longer F*d as it should be, however, unless F-mg is negligible (a slow lift). For F>>mg, the kinetic energy will dominate and throw off the calculation significantly.

Thus, for non-rigid bodies, only potential energy can be correctly determined from the center of mass alone. Kinetic energy analysis requires knowing the independent motion of all parts.

-Seth

wytcom : your r was indeed 0.5 and that is the value I used in the calculation. The 0.2 is a typo coming from 1/2.
The calculations are correct and we are not abusing physics. The energy for the work comes from the internal energy of the body. The body presses down on the earth with a force |F|. By Newton's 2nd Law the earth presses up on the body with a force|F|. The mechanics of a collection of particles (molecules) alows us to consider this force to be acting on the cm. The realization that this fact is so is a non-trivial physical insight that typically takes a page or two to rigorously prove and greatly simplifies many problems.

SethHoyt : The KE before is 0, the KE after is 0. KE does not enter into this problem.

aburr: well said.

Absolutely incorrect, aburr.....

It should be blatently obvious that if KE=0 the entire time, the object must be quasi-static. This is clearly NOT the case. There is no way to leave the ground and rise to some height without having a non-zero KE and momentum.

aburr's assertion is that all the internal energy expended to make the jump is included in the expression of average force times the distance the center of mass moves during the jump:  
Total energy = F*x

Then at the top of the jump all of the energy is in the potential energy of the jumper who is not moving at the top of the jump:
m*g*(h+x)

Thus we need not consider the kinetic energy during the jump at all:
F*x = m*g*(h+x)

I have been whining about the validity of using F*x as equivalent to the total internal energy expended in the jump.  But aburr assures me that this is ok.  It also works for the problem where the person is replaced by a massive spring in which case we know how to calculate the original internal energy.  So I am satisfied.

You are still not getting my point.

It must be agreed that the jumper does have KE>0 at liftoff, correct?

I proved that only the potential energy of the body *at liftoff* can be determined correctly by using the center of mass. You CANNOT find the potential energy at the apex this way, because the KE does not get included in the calculation of the total energy!!! Thus, your calculation is missing that energy, and you will get the wrong height. It simply does not work... please read the last example I gave about the weightlifter. I can't come up with a much simpler counterexample than that.

Please listen to what I am saying, I've studied physics for 6 years at the college level, and I do have some idea what I'm talking about.

I just want to clarify what I'm saying here...

The energy you are calculating by using F*d on the cm is only the correct total energy if the object is a rigid body or the motion is very slow (quasi-static). What we have here is a fast-moving non-rigid body. Thus, your calculated energy does not take into account the *internal motion* of the object. It does take some of the kinetic energy into account, but not all. The more internal motion the body has, the more your energy calculation is incorrect.

That is the "magic" that aburr (and myself!) has applied here to solve this problem.  Again, aburr's assertion is that F*d on the cm expresses the full initial internal energy of the jumper.  At the point when the jumper is just leaving the ground this total energy F*d is indeed split up between potential and kinetic.  But, so what?  It is this same total energy, F*d that is present at the top of the jump when all energy is only potential.  So we can skip the intermediate analysis and just use energy conservation at the top of the jump to solve the problem.

As I have been saying, it is not obvious that F*d (where d is the cm movement during the liftoff) is equivalent to the internal energy expended by the jumper during the jump.  But this is commonly used to solve this type of problem.  AND it gives the correct answer if we substitute the person with a spring where we can calculate the internal energy explicitly.  

I guess it remains to be proven that the use of F*d is correct here.  But it is not disproven by observing that there is both potential and kinetic energy at liftoff.

Comparison to spring:

Total energy:
E = (1/2)k*c^2      c is full crouch

Average force during jump:
<F> = k*c/2           Force goes linearly from k*c to 0

To determine which "d" to use, require that <F>*d = E:
(k*c/2)*d = (1/2)k*c^2       So
d = c          The full crouch!!!

Then for the jumper
<F>*c = m*g*(h + c/2)

So <F> = 61*9.8*(0.8 + 0.1)/0.2 = 2690 N

Conclusion:  If we insist on using F*d in the problem of the jumping person to represent the total energy, then we must use d = the full crouch.  This is necessary to capture the total energy of the jump.   This is tricky because THE GROUND DOES NO WORK ON THE JUMPER.

This is why I tried to clarify what I was saying. It's not that you don't take any kinetic energy into account by doing a rigid-body analysis, but you don't take any internal KE into account, only the cm kinetic energy. This is not correct for a non-rigid body. A correct analysis would be to use the total momentum, or to add on the internal KE to your total.

If you're confused about whether the ground not moving is a problem, it is not. One needs to realize that the most appropriate definition of force is dp/dt, and it simply represents the change in momentum of the two objects in contact. In this case, it is the earth and the jumper. In fact, the ground does move, we just don't care how much, and we could always determine that if we wanted to from what we know.

Take one more example of a cannon firing upwards and trapping the projectile to lift it.

If a cannon of mass M=100kg fires projectile of mass 10g upward with continuous force 10^4 N through a distance of 1m then catches it, how high does the cannon lift off the ground?

First, using the incorrect rigid-body analysis: Considering the center of mass of the cannon/projectile system, we calculate its motion to be about 0.1mm, through which a force of 10^4 N is applied (the ground feels this force). Using W=F*d, the work done on the system is 1 N*m. Equating with Mgh gives 1=980*h, thus h is about 1mm.

Now, correctly, doing what makes sense: Consider only motion of projectile, which is a rigid body. W = F*d = 10^4*1 = 10^4 N*m. Equating with Mgh yields 10^4=980*h, so h is about 10 meters now. This makes sense, because the cannon is pushing the projectile with 10 times its own weight for a meter before stopping it.

You should be able to see that applying the ground force through the cm's motion seriously underestimates the total energy of the system here. The internal energy cannot be detected from that information alone.

-Seth

BUT...

My spring analysis above is wrong   :-(

Executive summary: Work done by ground on person - Work done by gravity on person = 0
So F = mg(h + c/2)/(c/2) = 5380 N.

In my spring analysis above where I said to use F*d = E to determine which d to use, I was saying that the work done by the ground should equal all of the original internal energy provided for the jump.  But that is wrong.  Some of that internal energy goes into kinetic energy at lift off.

The correct spring analysis shows that the work done by the ground during take off of the spring is E/2.  This leads us back to using c/2 to calculate the work done by the ground on the jumper.  So I can now go back to my previous answer of 5380 N.

(This isn't as much fun as I thought it would be!)  

Ok, I'm going to break down and actually show you where your specific model fails. I tried using simple models to demonstrate the idea, but apparently it's not obvious how these extend more generally.

Here is the correct analysis of wytcom's continuous spring model:

We consider a spring that delivers constant force over the desired range. This can always be approximated with a particular choice of spring. We will distribute the mass equally along the spring.

At rest, the force in the spring as a function of y (distance from ground) is F(y)=(1 - y/h)*mg, which is just the weight above.

Applying a constant force F down from the top, we compress by c, to a new height of h-c. In doing so, we have increased the stored energy in the spring by F*c (work done). This is fine even if the body is non-rigid as long as we compress slowly.

Now, the spring has F*c more energy than is necessary to return to the rest position. Let us assume that the spring is constrained so that it cannot expand beyond a length h. This also explains why it can take more than zero force to compress.

Prior to releasing the spring, the force in the spring is F(y) = F + (1 - y/(h-c)) * mg. Note that the force on the ground is F+mg.

After the spring is released, it will release its entire stored energy. At the point of maximum extension, the spring will have a kinetic energy of F*c. Equating with mgh yields F*c=mgh, and solving for force: F=mgh/c

For the given values of m=61kg, h=0.8m and c=0.2m, F=(61)(9.8)(0.8)/(0.2) = 2400 N.

The force felt on the ground is F+mg = 2400 + (61)(9.8) = 3000 N.

As long as the object can deliver a constant force, and the cm moves by less than 0.2m during the jump, you will always get this answer. I had actually proved this using my momentum argument, but it seems to have fallen upon deaf ears.

-Seth

Seth's Cannon and projectile (hereinafter "ball") example (09/03/2003 12:53PM PDT) has got me thinking.

It graphically illustrates the difference between two of the methods.

However I believe (unlike Seth, unless I have misunderstood him) that the Cannon will rise by 1 mm, not 10 m.

Seth is assuming conservation of energy - and that all of the kinetic energy of the ball can be converted into potential energy (of the cannon and ball).

I think that there will be conservation of momentum as the cannon catches the ball.  As the weight of the moving object increases by 10^4 (from 10g to 100Kg) the velocity will decrease by 10^4 and thus the energy (=mv^2) will decrease by 10^4.  The other 99.99 % of the ball's energy will be transferred into heat and sound etc. by the catching device on the cannon.  This loss of 10^4 of the energy reduces the height the cannon will reach from 10 m by 10^-4 to 1 mm.

To take the general case of a non-rigid body (or a spinning rigid body) only that part of its kinetic energy which is represented by upward vertical movement of its cm (centre of mass) will get translated into potential energy.  

In the case of this question any other kinetic energy is undesirable as it increases the work / force required without increasing the jump height.  To get maximum momentum you require maximum vertical velocity of the cm.  Doing this with a steady force (to minimise the maximum force) requires maximum vertical movement of the cm (thus optimally exactly 0.2 m).

The only way I can see of reducing the force required is to allow some of it to be applied after the cm has reached its normal rest position (by using your heels to stand on your toes before "jumping").  If the question didn't limit the upward force to acting over 20 cm I would suggest that a 61Kg person with 80cm feet (or 80 cm shoes) could raise themselves 80 cm exerting a force of 61Kg (61 g N, 600 N).  

Is there a limit to shoe size in the high-jump ?

Ouch! You are correct, richardjb, I left out how the energy is to be conserved in the catching process. Conservation of momentum is the more correct analysis, which is what I had been saying anyway. I actually did think about the energy loss, but assumed there was a way to catch it without loss of energy. But the only way would be to have it incur elastic collisions, which would not transfer its energy. This comes straight from thermodynamics.

I should have just stuck to my original conservation of momentum argument...

-Seth

Hate to beat a dead horse, but this question is nearly identical to one from my step-son's Physics book.  He's a senior in high school and this problem was posed in a section previous to the sections dealing with energy, conservation of momentum and center of mass.  The only tools provided prior to this problem were the kinematic equations and force.  So, here is how I went about solving this one:

Step 1
Solve for time (t) to get to .8m in height using the kinematic equation: x2 = x1 + v1*t + 1/2a*t^2
If the max height was .8m, you can determine the total time it would take to fall by simplifying the above equation to this: y = 1/2g*t^2
The time to get to .8m is equal to the time it takes to fall, so .8m = -1/2gt^2
According to my calculations this comes out to t = 1.26 sec

Step 2
Solve for the initial velocity required to get the person to .8m.
I used the kinematic equation v2 = v1 + at
v2 = 0 because the final velocity is when the person reaches the peak of the jump
a = -g = -9.8 m/sec^2
t = 1.26 sec as computed in step 1
So, 0 = v1 + -9.8m/sec^2 * 1.26sec
With this calculation, I get v1 = 7.77m/sec

Step 3
Solve for the acceleration that occurred during the uncrouching.
I used the kinematic equation: v2^2 = v1^2 + 2a * (x2 – x1)
In this case, v2 is equal to v1 from step 2 because the final uncrouching velocity is equal to the initial leaving the ground velocity.
x2 – x1 = .2m because that is the distance that was crouched.  That is the distance in which the individual accelerates towards the velocity required to get the person to .8m off the ground.
v1 is equal to 0 because that is the initial velocity at the lowest point in the crouch.  The person then accelerates from v1 to v2 over a distance of .2m.
I calculated the acceleration to be: 150m/sec^2

Step 4
Now that you have the acceleration during the uncrouching, you can calculate the total force exerted on the ground by multiplying mass x acceleration:
F = ma = 61kg * 150m/sec^2 = 9150N