# Kinetic Energy with constant velocity

I am sorry for the question, but I am a noob in Physics. I don't understand why in the kinetic energy's formula the distance is out of the equation.

Kinetic Energy formula:

$$K = \frac{1}{2}m v^2 = n J$$

For example: if I run $5$ miles at $3m/s$, I spend the same calories or joules ($J$) if I run for few meters with the same velocity. It is not like in the real life. I don't understand why. Any hints?

• What you've written below the equation suggests that you're worried about why $distance$ doesn't appear in the equation. But above the equation you seem to be concerned about $mass$. Could you please clarify? Sep 2, 2018 at 18:03
• What is $n$ and $J$? Sep 2, 2018 at 18:10
• @garyp $n$ is a value and $J$ is Joule Sep 2, 2018 at 18:12

The kinetic energy is only part of the picture. If you were gliding in a perfect vacuum, your assessment would be right, it'd take the same amount of energy to travel different distances at the same speed (and it'd just take longer). (Although even here, you're body wouldn't be perfectly efficient in converting it's chemical potential energy into kinetic energy, so the total energy spent would be greater than the kinetic energy gained, the rest of the energy would go to heat.)

Running here on earth, you have loss mechanisms such as friction and air resistance. In addition, at a run each stride is like a small jump, so a lot of energy is probably spent fighting gravity. Running for a longer time means you need to counter-act these forces for longer which takes more energy. To be really realistic you'd need to take into account how these forces varied with your speed, and take into account your biological efficiency at different speeds, etc. Intuitively, this is also why bikes take much less energy to go farther, you're minimizing several of the loss mechanisms.

As you can probably tell, a proper calculation of all these effects is very difficult. Physics at this level is just trying to give you a baseline to begin these calculations and give you some intuition for the underlying principles.

As we I assume that( velocity =distance /time) internally they are related and we are moving means velocity is a measurable quantity. Not the distance taken as our prioty measurable quantity.

Consieder you running. If an observer watch you from a very high distance you can be considered as material point. Let's also assume that you start and finish firm ($v=0$), and that the start and the finishing line coincide. So we have: $$\Delta k=-\Delta U + W_{nc}$$ Where $W_{nc}$ is the work done by the non conservative force. $$\Delta k=\Delta U=0 \rightarrow W_{nc}=0$$

We can also write $W_{nc}=W_{1}+W_{2}$ where $W_{1}$ is the work done by your legs and $W_{2}$ is the work done by forces which oppose your motion. So: $$W_{1}+W_{2}=0$$ $W_{1}=F_{1}L$ and $W_{2}=F_{2}L$ If the forces are constant.

$$F_{1}=-F_{2}$$ Also $$P_{1}=F_{1}v$$ where $P_{1}$ is your power.

Now you must remember the first principle of thermodynamics for a system (closed): $$\Delta U= Q-W$$ Now, assume that $Q=0 \rightarrow \Delta U=-W$ $$\Delta U=-F_{1}L$$ $$\Delta U=-P_{1}\Delta t$$ NOw if you run for 3 minutes with a power of 1000 W your internal energy decreases of 180KJ. NOw if you run for 10 minutes with a power of 1000 W your internal energy decreases of 600KJ, and you feel more tired.

You are conflating work with energy, but there is a slight difference.

In fact: Work=change in Energy, i.e. U=ΔK.

Work is the accumulated force magnitude over distance along a path, so that an element of work dU=<F, ds>=F.ds.cos(θ) , where ds=elemental displacement along trajectory curve S , F=applied Force along S.

We integrate this and then call the upper and lower bounds "energy", or "kinetic energy" to be more specific, say K₁ and K₂, such that ΔK=K₂-K₁.

So, we are really taking about "energy levels", at specific positions on the path, not a real-time formula that is calculated at every position.

These "energy levels" are really potentials , we could call them "work potentials", similar to the idea of "field potentials" e.g."voltage".

In order for work to actively exist , a force that provides impetus to the mass must exist, therefore the energy is not "carried" by the mass unlike have you have been taught, but work (actually "average force x distance") is "instructed" into the structure of the mass via deformations which results in vibrations .

These internal vibrations (that do internal "work") reverb inside the mass until they dissipate to the exterior environment. That can only happen if there's exterior masses in contact with the object at positions along it's trajectory, at which point the "kinetic energy" (work potential) is "released"(force is transmitted) into said masses, so that even a mass that is moving at a constant velocity will have internal cyclical forces within that provide impetus and therefore inertia to the structure of the mass.

• To answer your specific example: Distance travelled S=5 miles, speed v=3m/s. Case 1: Not interacting with the ground (suspended above, neglecting air resistance). Your energy level is K=(½)M (3²)=4.5M Joules (if M is in kg). Case 2: Your foot impacts the ground and some of your momentum is transmitted to it resulting in an impulse, i.e. a force over time & a different energy level K'=(½)M ((3+Δv)²), where Δv=change in speed. Mar 9, 2021 at 0:19
• The work your body's mass M experiences with each impact (over distance ds) is: W=K'-K=ΔK=(½)M (6Δv+(Δv)²). Your initial speed was 3mps=3.28084 ydps. Say the stride length for is 2 strides per second, then your stride (Δs) is 3.28084 /2= 1.64042 yd. 5mi=8800yd=5364.48*Δs, so no. of impacts N=(#mi/Δs)(#Δs/ds)~ 5364.48*2~10729, total ΣW ~N.ΔK~(10729)(½)M(6Δv+(Δv)²)~5364.48M(6Δv+Δv²). However, Δv alternates from + to - as the ground decelerates you back, so pretty much all the 6Δv terms cancel leaving ΣW~5364.48M((Δv)²). The Δv is along the arc of a circle. Mar 9, 2021 at 1:08
• From circular motion,Δ(v')/Δt~ (v')²/2R, where v'=v'(foot rel pelvis). R is length of your lower leg. If your stride is 3.28084 yd, your total leg is roughly then 2R=3.28084 yd /Δθ. 40 to 60 degrees is about max that your ankle can turn down whilst in contact with the ground, & 15 degrees up. Say R=0.55yd, then Δθ~3.28084/(2*0.55)~0.902 rads~51.69 deg, & if v(foot rel knee)=0.5v', Δv/Δt~3.281²/(4*0.55)~(4.893 )ydps². 0.5s per stride over Δs=1.64042 yd, divided by2 steps =>Δv~1.223125 ydps. Thus: ΣW~5364.48M((1.223125 ydps/0.9144 yd/m)²)~(15.693)M kiloJoules~3.75 M kcal, where M=mass in kg. Mar 9, 2021 at 3:56
• Worth a look. You could think of your legs as compound pendulums. That's why I used v & v'. upload.wikimedia.org/wikipedia/commons/thumb/c/c9/… Mar 9, 2021 at 4:09
• To the Draft Science cult followers (a hermit guy on Youtube who doesn't believe in kinetic energy). KINETIC ENERGY for DUMMIES - VARIOUS EXPERIMENTS by STEPHEN BROUGH youtube.com/watch?v=YdwcRUamwh8 Mar 9, 2021 at 22:47