# If I jump off the ground is the net reaction force doing work on me?

I know this may seem a basic Physics question but I still have problems understanding the 'work' concept.

My muscles contract and they extend and push on the ground which in turn pushes on me . If I generate enough reaction force from the ground to overcome my 'weight' then it will accelerate my body off the ground. My body has experienced an increase in kinetic energy due to that 'Ground Reaction Force' therefore somehow it has done work on me? But the point of application between my feet and ground has not moved while that 'Ground Reaction Force' (impulse) has been applied , therefore no work has been done on me.

So this seems a bit of a paradox (to me) that the cause of my increase in kinetic energy is a ground reaction force but it hasn't done any work on me and vice versa (I haven't done any work on the earth).

Where am I going wrong?

• Re, "this seems a bit of a paradox" Maybe it would be less paradoxical if you consider that, in reality, you are not a rigid body. you have muscles and bones and joints between the bones; and during the entire span of time while your center-of-mass is accelerating upward, your feet are stationary, in contact with the ground. All of the interesting interactions during that time are taking place inside of you. Jan 8, 2021 at 14:59
• physics.stackexchange.com/a/598732/261792 see this answer here Jan 8, 2021 at 15:04
• An easy way to identify the force doing the work is looking at energy transfer. The earth is not losing any energy, so it can't possibly do work on you. On the other hand, you get tired from jumping, which means that your chemical potential energy decreases. Intuition suggests that since the transfer of energy (chemical to kinetic) is purely internal, the forces doing the work must also be internal. Sure enough, the forces doing the work are internal, spring-like muscular forces Jan 8, 2021 at 15:11
• @SolomonSlow Your comment is actually the beginning of an answer and contains an important idea: the human body is not rigid. Comments are for clarification/ simple hints/questions, not partial answers. Jan 8, 2021 at 17:20
• @BillN, Thanks, but I'm no physicist. My hobby is helping people to better understand the questions that they are asking. I'm not so good at actually answering them. Jan 8, 2021 at 17:27

I'm only answering so I can reminded everyone:

Including the human body will never provide insight into Newton's Laws.

So, instead, imagine you are a spherical mass ($$m$$), with one leg, and that leg is a coiled spring (constant $$k$$), contracted by length $$L$$

When you jump, the spring applies a force:

$$F = kx$$

through a distance $$L \ge x \ge 0$$

The ground doesn't do any work: it applied force through zero distance.

All the energy ($$\frac 1 2 kL^2$$) was in the spring, and each little element of the spring applied force to the adjacent piece through some distance.

Unlike muscles, you can actually work it our for a spring, which is a good exercise.

• But do you agree there is a force being applied to the contact point of that 'coiled spring leg' via the Earths surface? And that is the only external force acting on that spherical mass via that 'coiled spring leg'? But really it isn't being used to accelerate the mass , just constrain the movement of the coiled spring leg 'contact point' with the earth to allow it to build PE (from internal chemical energy) for internal transfer to KE? Jan 9, 2021 at 4:00

The earth didn't work on you, it just changed the direction of work done by your muscles on earth. Suppose you are using a manual pump to inflate tire of a bicycle now who does the work? You or the pump?

The work is obviously done by you, the pump just changes the the direction in which work is being done.

• Is your name purposely an anagram of Newton? Jan 8, 2021 at 16:51

The confusion related to work done by the normal force from the ground on the person, can be resolved by distinguishing between the motion of the center of mass of a system of particles, and the overall energy balance for the system of particles.

For a point particle, $$\vec F = m \vec a$$, and $$\int \vec F \cdot d \vec s =\Delta KE$$, where $$\vec F$$ is the net force on the particle, m is the mass of the particle, $$\vec a$$ is the acceleration of the particle, $$\vec s$$ is the displacement of the particle, and $$\Delta KE$$ is the change in kinetic energy of the particle.

As developed in physics mechanics texts, for a system of particles, the motion of the system can be considered as two separate motions: (1) that of the center of mass (CM), and (2) that of the particles in the system with respect to the CM. For motion of the CM, $$\vec F_{ext} = M \vec a_{CM}$$, and $$\int \vec F_{ext} \cdot d \vec s_{CM} = \Delta KE_{CM}$$, where $$\vec F_{ext}$$ is the net external force on the system, M is the total mass of the system, $$\vec a_{CM}$$ is the acceleration of the center of mass, $$\vec s_{CM}$$ is the displacement of the CM, and $$\Delta KE_{CM}$$ is the change in kinetic energy of the CM. That is, the motion of the CM is obtained by treating the CM as a point particle. Just as for a point particle, for the CM, $$\int \vec F_{ext} \cdot d \vec s_{CM} = \Delta KE_{CM}$$, but not all the external forces are associated with work using the general thermodynamics definition of work; the name pseudowork is sometimes used for such cases. Discussions of the concept of pseudowork can be found online. In such developments work is reserved to mean work in the thermodynamics sense, and that convention is used in the following discussion of jumping.

An example of pseudowork is the effect of the normal force from the ground on a person jumping. This is discussed in an article by Arons, available online. [Development of energy concepts in introductory physics courses: American Journal of Physics: Vol 67, No 12 , A.B Arons] The following discussion is from that reference. For the CM of the jumping person initially at rest, let M denote the mass of the person, $$\bar N$$ denote the average normal force of the ground on the person, and $$\Delta h_{CM}$$ the change in the difference of the height of the CM at the instant the feet leave the ground. For the CM treated as a point particle, $$(\bar N - mg) \Delta h_{CM} = {1 \over 2} m v_{CM \space f}^2$$ where $$v_{CM \space f}$$ is the velocity when the person just leaves the ground; $$-mg\Delta h_{CM}$$ is the work done by the force of gravity. $$\bar N\Delta h_{CM}$$ is not work in the thermodynamics sense because there is no relative displacement at the point where the normal force N is applied (unlike gravity, $$\bar N$$ does not act over a displacement distance), but in the context of the CM treated as a particle, $$\bar N$$ does pseudowork, because it changes the kinetic energy of the CM. Applying the first law of thermodynamics to the person as a closed system, $$Q + W = 0 = \Delta U + {1 \over 2} Mv_f^2 + mg \Delta h_{CM}$$, or $$-\Delta U = {1 \over 2} Mv_f^2 + mg \Delta h_{CM}$$, where the work by gravity is considered as the negative of the change in potential energy; $$\bar N$$ does no work. $$\Delta U = U_{final} - U_{initial}$$ is negative for this situation, and the internal energy of the person decreases to increase the KE and PE of the CM of the system. This decrease in internal energy is due to the “burning” of calories to provide energy for the jump. $$-\Delta U = \bar N\Delta h_{CM}$$; the decrease in internal energy from the energy point of view is equal to the pseudowork from the CM point particle point of view.

In conclusion, for the person jumping off the ground, the normal force on the person from the ground does no work, but from the viewpoint of the motion of the center of mass of the person the normal force can be considered to do pseudowork.

• Many thanks for this as pseudowork is something I have never heard of before. May 17, 2021 at 2:58
• I had never heard of pseudowork in any physics courses, but found out about it when puzzling over problems for a system of particles such as this problem. Glad this helps! May 17, 2021 at 12:58

When you push against the ground, you create a force dependent on your rate of change of momentum wrt time.

Gravity causes an opposing force due to the rate of change of energy wrt vertical distance.

When the two forces are equal (which can only be for an instant in time in this scenario), you will feel no force against the ground and will be effectively weightless. Accelerate beyond a rate of g and you will leave the ground.

Of course you are doing the work on yourself by using up a tiny bit of your body's mass to change the momentum of the rest of it.