# Can't wrap my head around the energy formula being half of mass times velocity squared [duplicate]

I'm trying to understand the "Why" of this equation but can't make sense of it, my intuition breaks down.

When I think intuitively of energy, I think of "How powerful the impact will be if an object collides with me". Let's say a 5kg object flying at 100kmh at me. I think of this as having a lot of energy, while 100 grams flying at 1kmh having little energy. I also think about how can I make use of it, maybe dropping the weight and compressing a spring to transfer it to something else, seems like mass and velocity is all I need.

Now, if I had to guess the formula of energy I would say that all that matters is the final velocity and the mass. I would imagine energy being m * v. But it's not.

Studying energy I keep encountering counter intuitive explantations and data that I feel like I don't need.

One being distance. Why would I care about the distance? If an 5kg object collides with me at 100kmh I don't even know where it comes from (or the distance it traveled) or how it accelerated... nor do I care, if I want to compress a spring with that energy it's irrelevant how it accelerated or the distance traveled. Why it makes a difference if the object was accelerated by gravity or teleported at 5 meters of my spring if both will have the same result on collision.

I get that If I have a 2kg object at 20m height and I drop it, to calculate the energy I have to make use of acceleration due to gravity. But for my brain, this is just to calculate the final velocity before impact, but it's not, as I need to divide mass by half as I take the average velocity, which seems to me is data that I don't need.

If someone can help me map this into my head and build intuitive thinking, don't want to end up just memorising and copy pasting the formula.

The classical "momentum" versus "energy" confusion !

Momentum is mass times velocity $$\overrightarrow{P} = m \cdot \overrightarrow{v}$$ and is a vector (points in the same direction as the velocity). Energy is a scalar and $$E = \frac{1}{2}m \cdot |v|^2$$

The common example of what this makes sense, is firing a rifle. Both the bullet and the rifle have equal but opposite momenta, so the overall momentum is preserved. This creates the recoil of the rifle. However, the bullet has way more kinetic energy otherwise the bullet could never inflict more damage than the rifle inflicts on you. That would make guns a lot less popular than they currently are.

In most cases energy quantities are the product of two linear (or field) quantities. Electrical power is voltage times current. Acoustic intensity is pressure times particle velocity. In a mass spring system, power is force times velocity. That's a little less obvious here since the kinetic energy is proportional to the product of the velocity with itself. Since the velocity is a vector and the energy isn't the specific product here is the dot product, i.e.

$$E = \frac{1}{2}m \cdot \overrightarrow{v} \cdot \overrightarrow{v}$$

There are different ways to explain it but that's already been done so I recommend googling "momentum vs kinetic energy" and finding the explanation that fits your intuition the best.

I will try to avoid too much calculus. Suppose that an object of mass $$m$$ is moving in one dimension under the influence of a force $$F$$. In a very small time interval $$\Delta t$$, the impulse is $$F\Delta t$$. In the same time interval, the object moves a small distance $$\Delta x$$. The work over that interval is then $$F \Delta x$$. If the object is moving at a constant (approximately, because $$\Delta t$$ is small) speed $$v$$ in this interval, then using $$distance = rate*time$$ we can relate the two using $$\Delta x = v\Delta t$$. Therefore the work is $$Fv\Delta t$$.
Newton's second law gives us $$F=ma$$, and over our small time interval, $$a \Delta t = \Delta v$$. Therefore the impulse is $$\Delta p = ma \Delta t = m\Delta v$$, which is proportional to the change in velocity. We can make the same argument for the work. The work done in the interval is $$\Delta W = Fv\Delta t = ma\Delta t \Delta v = mv\Delta v$$. By an application of the chain rule, $$\Delta (\frac{1}{2}v^2) = v\Delta v$$, and so $$\Delta W = \frac{1}{2}m\Delta(v^2)$$, which is proportional to the change in the square of the velocity.