# How Can We Jump When the Ground Does No Work?

Common Topics: spring, energy, mass, work, force

It is relatively common on Physics Forums to see arguments that are effectively similar to the following:

When we jump off the ground, the ground does not move. Because of this, the force from the ground on us does zero total work. Since the force does no work, we cannot gain any kinetic energy. We therefore cannot jump off the ground.

Now, the conclusion here is obviously false. The world high jump record is 2.45 meters, definitely larger than zero. So where did the energy come from? This Insight seeks to clarify this in a fairly accessible way.

## An idealized example

Before jumping off into bipedal mammals competing in the high jump, let us look at an idealized example. This example will help us understand what is going on a little better.

A mass ##m## has an ideal spring of length ##\ell## and spring constant ##k## attached to it. The mass and spring are pressed against a fixed wall such that the spring is compressed by a distance ##D##, see the figure below. A mass ##m## attached to a spring of length ##\ell## next to a wall. When pressed towards the wall, the spring compresses by a distance ##D##.

In other words, the mass of the spring is zero, and the force at its ends is given by Hooke’s law. All of this occurs in a horizontal plane, meaning that we do not need to deal with gravity.

Once released, the spring pushes the mass away from the wall. Similar to the jump off the ground, the wall provides no work. By the same reasoning as in our example argument, the mass cannot move away from the wall.

## Where is the energy?

So where does the energy come from? Because the spring will certainly launch the mass away from the wall. In order to answer this, let us first look at the process of compressing the spring. In particular, we consider a small segment of the spring between the coordinates ##x_0## and ##x_0 + \Delta x## when the spring is relaxed. The compressing force pushing on its ends is ##F = -\kappa \epsilon## in accordance with Hooke’s law (see the figure below). Here ##\epsilon## is the strain and ##\kappa = k \ell##. To compress a spring segment ##\Delta x##, forces equal in magnitude but opposite in direction act on both ends. The displacement at the top of the segment (red) is larger than that at the bottom (blue). Therefore, the upper force does positive work of a larger magnitude than the negative work of the lower force.

Changing the strain by ##d\epsilon##, the lower end of the string segment moves by ##x_0 d\epsilon## and the upper by ##(x_0+\Delta x)d\epsilon##. The total work done on the segment becomes $$dW = F x_0 d\epsilon – F (x_0+\Delta x) d\epsilon = \kappa \epsilon \Delta x \, d\epsilon.$$ Integrating this from no strain to a strain ##\epsilon_0## leads to $$W = \kappa \Delta x\int_0^{\epsilon_0} \epsilon\,d\epsilon = \frac{\kappa\epsilon_0^2}{2} \Delta x.$$ This is the total energy stored in the spring segment at strain ##\epsilon_0##.

That the total energy stored in the spring is $$W = \frac{kd^2}{2}$$, where ##d## is the compression of the spring and ##k = \kappa/\ell## is the spring constant, follows directly from the above.

## Energy flux

The discussion above suggests the idea that energy can enter or exit an object and remain as internal energy. This occurs through forces acting on the object performing work. A force ##\vec F## acting on an object over a displacement ##d\vec r## will do a total work of ##\vec F \cdot d\vec r##. In the example above, ##x_0 d\epsilon## replaces ##\vec dr## for the lower end as this is the lower end’s displacement and we work in one dimension. Similarly, we have a directed one-dimensional force ##F## instead of the three-dimensional vector ##\vec F##.

The quantity ##F x_0 d\epsilon = – \kappa \epsilon x_0 d\epsilon## is, therefore, a measure of the amount of energy flowing upward through the spring at position ##x_0## when the strain changes by ##d\epsilon##. When ##\epsilon## is negative, i.e., when the spring is compressed, energy will flow upward if ##d\epsilon## is positive. In other words, when the spring is compressing energy flows left in the spring and right when it is decompressing.

## Launching the mass

The spring will decompress during the launch of the mass. The internal energy stored in the spring then flows from the spring into the mass. Denoting the compression of the spring ##D(t)##, we find that $$D(t) = D_0 \cos(\omega t)$$ with ##\omega^2 = \kappa/m\ell## during the launch, where the initial compression is ##D_0##.

The launch time interval is ##0 \leq t \leq \pi/2\omega##. The strain ##\epsilon## is related to ##D## as ##\epsilon = -D/\ell##. We therefore obtain $$\frac{d\epsilon}{dt} = -\frac{D'(t)}{\ell} = \frac{D_0\omega \sin(\omega t)}{\ell}.$$

Consequently, the energy flowing up through the spring at position ##x## is $$\frac{dW}{dt} = -\kappa \epsilon x\, d\epsilon = \kappa \frac{D_0\cos(\omega t)}{\ell} x \frac{D_0\omega \sin(\omega t)}{\ell} = \frac{\kappa D_0^2}{2\ell^2} \omega x\sin(2\omega t).$$

It is natural that this grows linearly with ##x##. As all energy released from the spring flows into the mass, the energy flow gets larger the closer to the mass we get.

## Relation to the jumper

A jumper’s legs are by no means an ideal spring. However, the discussion above does give some insight into the issue presented in the beginning:

• The upper body will receive net work from the legs much like the mass received net work from the spring during launch.
• The net work from the ground is zero.
• The energy is provided from internal energy stored in the jumper’s muscles. Just as the energy here was provided from internal energy stored in the spring.

Some differences are also notable:

• Unlike the spring, the jumper’s lower body will have non-zero kinetic energy at the end.
• Energy will also be lost in the form of heat as the efficiency of conversion of internal energy to macroscopic kinetic energy is not 100%.

While the ground does not do work on the jumper, the jumper’s momentum is provided by the force from the ground. This momentum is distributed throughout the jumper’s body by internal forces.

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20 replies
1. Orodruin says:
The force results in an equal in magnitude but opposite in direction change in momentum of jumper and earth away from the center mass of earth and jumper. Due to the earth being massive, the change in speed of the earth is extremely tiny, so almost all of energy goes into the jumper.

See #20-#22. The misconception that is being addressed is that the ground not doing work on the jumper would prevent kinetic energy buildup. This misconception is solved by showing how internal energy can be released by parts of the body moving at different velocities. Due to the small motion of the Earth (neglected in the Insight), the energy flow into the Earth is small (and amounts to the ground doing negative work on the jumper).

2. Orodruin says:
The insight mentions the jumper’s lower body doesn’t move,

No it doesn’t. In fact, the point being made is that the lower body extending (although in the example idealised by a spring) is what gives an energy flow into the body from the energy release in the lower body.

Edit: In particular, the Insight explicity states
“Unlike the spring, the jumper’s lower body will have non-zero kinetic energy at the end.”
The zero energy of the spring results from the spring in the example being an idealised massless spring. This is in order to simplify the considerations but does not affect the main point.

3. Orodruin says:
IIRC, enthusiastic fans jumping up and down in unison at stadia routinely register on nearby seismo’ sensors…

Sure, but as mentioned above that transfer is generally into the ground because the floor is not entirely rigid. I also strongly suspect that most of the energy transfer occurs on the landing *thump* rather than at the jumpoff.

( Don’t forget the thunderous ‘Thumper’ trucks used for geo-surveys… )

We have also all seen Dune … Thumpers can be very useful when dealing with enormous sandworms. 4. Orodruin says:
So yes, as @jbriggs444 said we are making the approximation that the Earth is stationary. However, since the Earth is pushed down, the miniscule energy that is transferred across the jumper-Earth contact area would actually flow from the jumper and into the Earth so this would make the misconception even more puzzling. The same analysis applies however and the energy released in the jump starts as internal energy in the jumper’s muscles.

Edit: We are also completely sweeping the entire issue of using other inertial frames under the carpet. Opening that is an entirely different can of worms…

5. jbriggs444 says:
in classical mechanics, doesn’t the earth technically "move"? Equal and opposite sort of thing? It’s just that the mass of Earth is so huge in relation to the human that it’s negligible?

Yes, you are right. We are agreeing to neglect that.

6. Orodruin says:
Another reason to deal with idealisation is to be able to highlight certain aspects of a problem without running into unnecessary complications. This is the case here. What is being discussed is how a force from the ground can lead to an object moving without providing any work within the context of classical mechanics. The example highlights this.
7. Orodruin says:
Yes. Unlike mathematics, a big part of physics is about making descriptions of real situations that may not be completely accurate but are accurate enough to describe what you want at a sufficiently precise level.
8. Orodruin says:
Effectively it is a description of how stored energy can be released without the ground doing any work. To use an idealised spring is a way to be able to focus on how the energy flows in the system without the complications of needing to also discuss the kinetic energy in the launch mechanism. Similarities and differences in the case of a jump are discussed at the end.
9. Delta2 says:
Btw @Orodruin implies that our legs play the role of the spring as we first bend them and lower our self down (compression of spring) and then we de -bend them (sorry my english fail here cant find the proper word) in order to jump high.
10. Orodruin says:
It is what we call an idealised spring. It does not exist in reality. It is a way to be able to neglect certain aspects in order to focus on the relevant issues in a physics problem while still getting a reasonably accurate result for many situations where you have a spring that is close to being ideal. An ideal spring has no mass and responds to compression and elongation with a linear force. A real life spring does neither of those except approximately.
11. Orodruin says:
If the spring is not attached to the wall, there is no external force to the spring+mass system so its momentum is conserved.

It does not need to be physically fixed to the wall to compress the spring.

12. Orodruin says:
Wouldn’t the explanation be simpler with a spring that is not attached to the wall (so no oscillation)?

The spring is not attached to the wall for this very reason. I might have made it clearer in the first image though. Also note that I have been careful to only discuss the launch up to the time when the spring is back to its unloaded length.

13. Delta2 says:
The spring is not being de-coupled from the wall entirely. It is still prevented from interpenetrating. It can push. But it is detached so that it cannot pull. It is not glued, nailed, screwed or otherwise fastened down.

Ok I see it can push but not pull, in that case momentum is not conserved and yes we will not have oscillation.

14. jbriggs444 says:
If the spring is not attached to the wall, there is no external force to the spring+mass system so its momentum is conserved.

The spring is not being de-coupled from the wall entirely. It is still prevented from interpenetrating. It can push. But it is detached so that it cannot pull. It is not glued, nailed, screwed or otherwise fastened down.

15. Delta2 says:
Sorry, but I don’t see how the spring being attached or not to the wall changes anything in this respect.

If the spring is not attached to the wall, there is no external force to the spring+mass system so its momentum is conserved.

16. DrClaude says:
Nope. From conservation of momentum, the body cant attain substantial momentum because the spring has negligible mass hence it has negligible momentum after the decompression happens.

Sorry, but I don’t see how the spring being attached or not to the wall changes anything in this respect.

17. Delta2 says:
Wouldn’t the explanation be simpler with a spring that is not attached to the wall (so no oscillation)?

Nope. From conservation of momentum, the body cant attain substantial momentum because the spring has negligible mass hence it has negligible momentum after the decompression happens.

18. DrClaude says:
Wouldn’t the explanation be simpler with a spring that is not attached to the wall (so no oscillation)?
19. Delta2 says:
This is one of the "rare" (which is not so rare after all but it is rare regarding to what is being taught and problems solved in introductory physics) cases in physics where the force that changes the momentum is not the same with the force that changes the kinetic energy. In this example the force from the ground changes the momentum , but internal forces in the knee and in the torso change the kinetic energy.
20. Orodruin says:
You start with a horizontal spring but in the analysis you refer to "lower end" and energy flowing upward. It may be confusing.