Can you derive the formula for Kinetic Energy?

Remember, this derivation is examinable and needs revising : )

[pabloreyes]

Great explanation. I have one question though.
The derivation of the formula for kinetic energy stems from the third kinematic equation. But if the kinematic equations are only valid for uniformly accelerated motion, does the formula for kinetic energy still apply for objects with non-uniform acceleration? If so, how do we derive it then?

[BTWPhysics]

The equation KE = 1/2mv^2 for kinetic energy is incorrect because the other equations Wnet = ΔKE and Wnet = fnet * d you're really referring to that are used to derive the equation for kinetic energy are incorrect. We can first take a look at the inconsistency here by comparing these two equations for net work and net force.

Wnet = Fnet ⋅ d
Fnet = ma

We know that a net force causes an acceleration and we also know that net work can only be done by a net force if it causes a displacement. We can see that "net force" and acceleration are on opposite sides of the equation for net force, but we find net force and displacement on the same side of the other equation for net work. So, why aren't net force and displacement on opposite sides of the equation since a net force causes a displacement in the same way that net force and acceleration are on opposite sides of the equation and a net force causes an acceleration? If we consider the equation for net force to be correct, then we would expect to see consistency with the equation for net work and find net force and displacement on opposite sides of the equation. Let's take a look at the net work equation with deriving the equation for kinetic energy like shown in the video.

Wnet = Fnet * d
Wnet = ma * d
Wnet = m * ad

a = (V - U)/t
d = 1/2( U + V )t

U = 0

a = V/t
d = 1/2Vt

ad = 1/2vt * v/t
ad = vt/2 * v/t
ad = v^2t/2t
ad = v^2/2
V^2 = 2ad

ΔKE = m ⋅ ad
ΔKE = m ⋅ V^2/2
ΔKE = m V^2/2

We can see that the equation Wnet = Fnet ⋅ d causes us to multiply two kinematic equations, thereby, deriving a third kinematic equation V^2 = 2ad shown in the video which has time factored out. We can refer to Occam's razor which, basically, states that entities should not be multiplied unnecessarily. It is not necessary to multiply two kinematic equations in order to derive a third kinematic equation with time omitted. We can show this using the other kinematic equations.

a = V - U/t
d = 1/2 at^2 + Ut

U = 0

a = V /t
d = 1/2 at^2

2d = at^2
2d/a = t^2

t = √2d/a
at = V

a * √2d/a = V
V = √2ad

We can see that we have derived the equation V = √2ad through substitution. We can refer to Occam's razor again which tells us that when we have two competing theories or explanations for something that the simplest of the two is preferred to one that is more complex. In this case, we have two competing equations that have time factored out. The simplest equation of the two is V = √2ad. We did not have to multiply two kinematic equations unnecessarily like with the equation V^2 = 2ad and there also was no need to square the velocities.

It is easy to believe that work is a function of displacement because this aligns with our everyday experience. We see that we had moved an object a certain distance and we also feel like we did work when we expend energy. The same goes with lifting an object up against gravity. But, ask yourself, is displacement the only possible factor other than net force to determine the net work? We focus on the distance we moved an object, but we lose track of time. It is easy to lose track of the time that passes when doing work and, so, we don't factor time in with how much work is done. Yes, the displacement increases with work, but so does the time increase. Let's take a look at the two possible factors starting with the net work equation.

a = V/t
d = 1/2Vt

Wnet = Fnet * d
Wnet = ma * d
Wnet = m * ad
Wnet = m * (V /t) * 1/2Vt

We can see that the equation Wnet = Fnet ⋅ d causes us to multiply two kinematic equations like shown before. Let's now take a different approach using time as the factor.

d = 1/2 ⋅at^2 + Ut

U = 0

d = 1/2 ⋅at^2

md = m * 1/2(at^2)
md = 1/2 * mat^2
md = 1/2 * Fnet * t^2
Wnet = 1/2 * Fnet * t^2

We can refer to Occam's razor which basically states that entities should not be multiplied unnecessarily. It is not necessary to multiply two kinematic equations considering that we naturally get net force as shown and that we have time which is a possible factor that increases with work done. Occam's razor tells us that when we have two competing theories or explanations for something that the simplest of the two is preferred to one that is more complex. In this case we have two competing factors of displacement and time and two competing equations for net work. The equation Wnet = 1/2 * Fnet * t^2 is the simplest out of the two equations and, hence, time is the correct factor. We can see that we did not multiply two kinematic equations with this approach and that we naturally get Fnet. Sure, we had to multiply by mass, but that is in both cases. We can see that with the product md that displacement increases with the independent factor of time. So, this equation has net work as a function of time as opposed to displacement.

Wnet(md) = 1/2 ⋅ Fnet ⋅ t^2
Fnet = ma

Let's now take it back to the inconsistency that I have pointed out before. We now compare the two equations for net work and net force shown above. We can see that the displacement that is caused by the net force is on the opposite side in the net work equation. This is now consistent with acceleration that is caused by a net force which is on the opposite side in Newton's equation. There is no more inconsistency.

We can also see that the unit for net work is kilogram-meter, as reflected in the product md, which is not a unit of energy. This means that the net work done is not equal to the change in kinetic energy and the equation Wnet = ΔE is incorrect. This also means that the equation KE = 1/2mv^2 for kinetic energy is incorrect.

[oogaboogaabrar]

for deriving kinetic energy, when i got to work done = mass x acceleration x displacement, i rewrote acceleration as change in velocity/change in time, and led me to get kinetic energy as just mv^2. can you explain please why this doesnt work?

[elysonpanolino]

now derive the formula for work
It seems more fundamental

why does W=Fd

[grahamflowers]

There is no kinetic energy in a moving mass there is force Mv squared kinetic energy is the energy of consistent work from a consistent force regards Graham Flowers MEng

[ruanholtzhausen]

Well W=∆KE but it's fine cus it aint KE it's ∆KE.