How can the kinetic energy equation be intuitively understood?

In short, momentum is vector and kinetic energy is scalar.

$$\begin{array}{}\text{(1)}& \overrightarrow{p}=m\overrightarrow{v}\end{array}$$ $$\begin{array}{}\text{(2)}& T=\frac{1}{2}m{\overrightarrow{v}}^2=\frac{\overrightarrow{p}}{2m}\end{array}$$

Two momenta in opposite direction is $0$. Total energy and momentum both are conserved, let total energy =kinetic energy + potential energy. Here kinetic energy changes its form (it mostly converts to other energy since energy can never be destroyed) while momentum is conserved body to body. In equation (2) I had shown that kinetic energy is mostly relevant to momentum (but they aren't same), the core difference here is one is scalar another is vector. Let's look at their units (other prefers dimension but I like units), units of momentum is $kgm$ while units of kinetic energy is $kg{m}^2$. Kinetic energy is movement energy of an object. Think that, in a system some kind energy is converted to kinetic energy, so the energy will make particles move in that system.

If two objects move at the same speed then the more massive one has more quantity of motion and if two objects have same mass then the faster one has more quantity of motion.

Kinetic energy is energy of moving object.

An integral form might be more helpful to understand the difference,

$$\overrightarrow{p}=\int \overrightarrow{F}\mathrm{dt}$$ $$W=T=\int \overrightarrow{F}\cdot \mathrm{d}\overrightarrow{\mathrm{s}}$$

Further reading :

1. http://www.differencebetween.info/difference-between-kinetic-energy-and-momentum
2. https://physics.stackexchange.com/q/16160/315444

Your analysis seems pretty good. To take an equivalent but more traditional example, imagine we throw a ball upwards. Ignoring air resistance and approximating the gravitational force as constant, the ball will accelerate downward with constant acceleration. If the initial (upwards) velocity is $v_0$ and the downward acceleration is $a=-g$, the ball will reach the apex at time $t=v_0/a$ at which point it will have zero velocity and thus zero kinetic energy. It will have traveled a distance of $v_0^2/(2a)$. In general, work is force applied over a distance. In this case, the work done by the gravitational field is the (constant) force, $F=-mg$, applied over this distance, $v_0^2/(2a)$, giving a work done of $m{v}_0^2/2$.

Moving more towards your analysis, once we fix a constant force (and constant mass), then we know work is directly proportional to the distance, $d$, the mass travels, i.e. $W=Fd$. Your analysis shows that $d$ is quadratic in the initial velocity, thus work is quadratic in the initial velocity. I believe this was your goal. The slightly more detailed analysis in the previous paragraph will reproduce exactly the kinetic energy expression if you wanted more.

posted over 3 years ago

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3y ago

Derek Elkins‭

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Change in momentum is force times time. On the other hand, change in kinetic energy is force times distance (more accurately, the component of the force along the movement times distance; a force perpendicular to the movement doesn't change the energy).

Now imagine you apply a force for a given time to an object initially at rest. The momentum will grow by the product of the force and the time. Since the object will have moved a certain distance after that time (and certainly in the direction of the force), its energy will also have increased by a certain amount. Also since a constant force means a constant acceleration, the velocity will also have increased by a certain amount.

Now imagine that you apply the same force for another second. For the momentum it's simple: The product of force and time is, of course, the same again, and therefore the momentum has grown by the same amount. Similarly, the acceleration was the same, and therefore the velocity has grown by the same amount. It is also easy to check that the same applies for a third acceleration phase, a fourth acceleration phase, and so on. Thus the growth of velocity and the growth of momentum are proportional.

On the other hand, if we look at the change of energy, we see that for the second acceleration, we get a greater energy growth, as the object was already moving, and therefore the total covered distance is larger.

So how much larger is the energy growth? Well, since the force was the same, it is the same as the difference of the covered distance. Now we had a constant force, and therefore uniform acceleration, which means the average velocity during the acceleration is just the arithmetic mean between the initial and final velocity.

So if we denote the final velocity of the first acceleration with $v$, then, because the initial velocity was zero, the average velocity during the first acceleration phase was $(0+v)/2=v/2$, and therefore the covered distance was $vt/2$, where $t$ is the time of acceleration.

On the other hand, the average velocity of the second acceleration phase is $(v+2v)/2=3v/2$, and thus the covered distance is $3vt/2$, o three times the distance of the first acceleration phase. Thus the total change of energy (and thus the total kinetic energy, since we started out with an object at rest, thus without any kinetic energy) is four times the kinetic energy after the first acceleration. Thus at twice the speed we have four times the kinetic energy.

If we do a third acceleration phase, the average now is $(2v+3v)/2=5v/2$, thus the kinetic energy at three times the speed gets a factor $1+3+5=9=3^2$. Similarly, at four times the speed, the kinetic energy gets a factor $1+3+5+7=16=4^2$. So we see that the kinetic energy goes with the square of the velocity.

This is related to the fact that the covered distance of an uniformly accelerated object is proportional to the square of the time of acceleration. Indeed, with constant force starting from rest, velocity is proportional to time, while covered distance is proportional to the square of time, thus covered distance (and therefore kinetic energy) is proportional to the square of velocity.

posted over 3 years ago

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celtschk‭

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