# What is the force behind this punch? [closed]

I originally asked this in worldbuilding due to the context, though I suppose it's just a physics question.

I was never great at physics in school but let's say there's a super human and they're flying at around 450 km/h and they weigh around 95 kg. They have their fist stretched out in front of them. How much force is behind this fist? If it punched someone how much weight would it feel like if that's how it works? I know impact time has something to do with it all but I'm not quite sure what.

I used some online calculator and got 742188 Joules but I'm not quite sure what this means, to be honest. (I know now this is regarding energy and not force or momentum).

For simplicity's sake let's say the target is a pillar that is stationary and not fixed to the ground. It's 2m tall and weighs 100kg while the hero is flying parallel to the ground.

## closed as unclear what you're asking by sammy gerbil, Jon Custer, John Rennie, Kyle Kanos, glSAug 3 '18 at 12:51

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Force is an interaction property, not a system property. There is no force until impact and then it depends on how what they hit behaves (how massive it is, if it is fixed down, in contact with a surface or freely flying, how it was moving prior to impact, how much the material gives, and so on...). Your online calculator gave you a kinetic energy for your hero, which is not force. Long story short: the question in its current state is meaningless as physics and the colloquial meaning that you intend is very complicated to address. – dmckee Jul 29 '18 at 3:43
• Right. So I know I need to factor in a lot of things like impact time and things like that. I'll edit my question. Thank you. – quid Jul 29 '18 at 4:51
• Joules is also not the units for force. Joules is the unit for energy. – QuIcKmAtHs Jul 29 '18 at 5:04
• Possible duplicate of How do I describe and calculate the effect of an impacting object? – sammy gerbil Jul 30 '18 at 16:49
• – sammy gerbil Jul 30 '18 at 16:57

## 1 Answer

With some assumptions, this might be solved in a classical "high school way".

First, convert the given speed to the basic unit $\frac{\text{m}}{\text{s}}$:

$v_{\text{human}} = 450 \frac{\text{km}}{\text{h}} = 125 \frac{\text{m}}{\text{s}}$ (to change the units in this direction, you have to divide by 3.6).

Then assuming a perfect inelastic collsion (i.e. human and column continue moving together as one object; the human pushes the column forward while the column stays in contact with his fist), the speed of the combined moving mass is calculated using conservation of momentum as

$$v_\text{both} = \frac{m_{\text{human}} \cdot v_\text{human}}{ m_\text{human} + m_\text{column}} \approx 61\frac{\text{m}}{\text{s}}$$

This means that the column has to be accelerated to this speed very quickly during the impact. The column will have kinetic energy

$$KE_\text{column} = \frac{1}{2} m_\text{column}v_\text{both}^2 \approx 186 \text{kJ}$$ at this speed.

Now the force can be calculated as the change in (kinetic) energy over a certain distance. Let's assume $10 \text{cm}$ for this distance where the acceleration happens:

$$F = \frac{\Delta KE}{\Delta s} \approx \frac{186 \text{kJ}}{0.1 \text{m}} = 1.86 \text{MN}$$

So assuming perfect inelastic collision and that both objects continue moving as a whole and estimating the deformation of the hero's arm to be 10 cm, we arrive at 1.86 mega newtons which is the [t]hrust of Space Shuttle Main Engine at lift-off.