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I don't understand why the work represent the change in energy of a system. I understand the work energy theorem but where did we get the formula of the work from in the first place. Also when I search for a derivation of the kinetic energy the only thing I find is a derivation using the work formula. Isn't there any other way to derive it? I know that work and kinetic energy can be derived from one another but which formula came first and how?? Sorry for my bad English.


marked as duplicate by ZeroTheHero, M. Enns, user191954, Buzz, Jon Custer Jan 21 at 3:43

This question was marked as an exact duplicate of an existing question.

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    $\begingroup$ This is a similar question. $\endgroup$ – cal Jan 20 at 11:12

What we want to show here is that the work done on the object, as given by the equation F*s is equivalent to the change in kinetic energy of the object.

Consider an object of mass M. Suppose we push the object with a constant force, F for duration T seconds.

v1: Initial velocity

dv: Change in velocity after time T

The displacement of the object is given by: v1 * t + 0.5(f*t^2/m). Hence, according to our definition of work, work done is: f * (v1 * t + 0.5(f * t^2/m))

Now we express the work done using the work energy theorem:

Final KE = 0.5 *m * (v1 + dv)^2 = 0.5 * m * (v1 + ft/m)^2 = 0.5 * m * (v1^2 + 2(v1 * f * t)/m + (f^2 * t^2)/m^2) = 0.5 * m * v1^2 + (v1 * f * t) + 0.5 * (f^2 * t^2)/m

Initial KE = 0.5 * m * v1^2

Change in KE (Work done) = Final KE - Initial KE = (v1 * f * t)/m + 0.5 * (f^2 * t^2)/m = f * (v1 * t + 0.5 * (f * t^2/m)).

Hence, we can see that the work equation and work energy theorem are equivalent expressions.

We can use a 3D curve to visualise this relationship: https://www.geogebra.org/3d?lang=en. Let the y axis be T, x axis be displacement and z axis be KE. Plot in the equations: x=a * y + 0.5(f * y^2/m) (Displacement against time) and z= a * f * y + 0.5(f^2 * y^2/m) (KE against time). a, f and m are constants that you can assign. If you look at the projection of the curve on the xy plane, it is probably linear. This is the graph of F*s, as derived using the relationship between displacement and time and KE and time.


Here is one example of how to go from a kinematic equation to kinetic energy:

An object that is dropped from rest, accelerates under the influence of gravity, and gains velocity as it loses height. This means that gravity is applying a force to that object, and since the object’s displacement is parallel to the force of gravity, gravity is doing work on the object. This leads to the following kinematic equation:

$v_f^2 = v_i^2 + 2g\Delta y$ (1)

where “g” is the acceleration due to gravity, delta-y indicates displacement in the vertical direction, and “down” is taken as the positive direction. Since the initial velocity is zero (the object was dropped from rest),

$v_f^2 = 2g\Delta y$ (2)

$v_f^2/2 = g\Delta y$ (3)

Multiplication of both sides of the above equation by the object’s mass yields

$mv_f^2/2 = mg\Delta y$ (4)

The left hand side of equation (4) is the kinetic energy of the falling object (energy of motion), and the right hand side of equation (4) is the gravitational potential energy of the falling object. Equation (4) states that as energy is conserved, the falling object loses gravitational potential energy, and it gains an equal amount of kinetic energy as it does so.


Energy of the system is the quantity that is conserved as a result of invariance with respect to translations in time, i.e. the laws that govern your system are the same irrespective of the time (the actual behaviour may depend on time, given suitable initial conditions). I like this definition since it focuses on what is important, not some magic formula, but the general (usually desired) property of the system.

What follows relies on Lagrangian mechanics. In case you are lost, I suggest looking into Goldstein's "Classical Mechanics"

Kinetic energy is the type of energy that a 'free' particle would have. So let's look at the Lagrangian for a free (non-relativistic) particle with mass $m$: $L=\frac{1}{2}m\left(\frac{d\mathbf{r}}{dt}\right)^2$, where $\mathbf{r}$ is the position of the particle, and $t$ is time.

For a general Lagrangian $L=L\left(t,\,\mathbf{r},\,\frac{d\mathbf{r}}{dt}\right)$:

$\left(\frac{\partial L}{\partial t}\right)_{\mathbf{r},\,\frac{d\mathbf{r}}{dt}}=\frac{dL}{dt}-\frac{d\mathbf{r}}{dt}.\left(\frac{\partial L}{\partial \mathbf{r}}\right)_{t,\frac{d\mathbf{r}}{dt}}-\frac{d^2\mathbf{r}}{dt^2}.\left(\frac{\partial L}{\partial\left( d\mathbf{r}/dt\right)}\right)_{t,\mathbf{r}}=\frac{d}{dt}\left(L-\frac{d \mathbf{r}}{dt}.\left(\frac{\partial L}{\partial \left(d\mathbf{r}/dt\right)}\right)_{t,\mathbf{r}}\right)$

The second step requires use of E-L equations. Thus quantity in the brackets on the right is conserved if $\left(\frac{\partial L}{\partial t}\right)_{\mathbf{r},\,\frac{d\mathbf{r}}{dt}}=0$, i.e. if system does not explicitly depend on the time (and is thus invariant under time-translations).

This is true for the free-particle Lagrangian, where the conserved quantity is:

$L-\frac{d \mathbf{r}}{dt}.\left(\frac{\partial L}{\partial \left(d\mathbf{r}/dt\right)}\right)_{t,\mathbf{r}}=\frac{1}{2}m\left(\frac{d\mathbf{r}}{dt}\right)^2-\frac{d\mathbf{r}}{dt}.m\frac{d\mathbf{r}}{dt}=-\frac{1}{2}m\left(\frac{d\mathbf{r}}{dt}\right)^2$

which is the energy of the system (the sign is due to convention). So let the energy of the free-particle be $U=\frac{1}{2}m\left(\frac{d\mathbf{r}}{dt}\right)^2$. If the energy is changing the change in energy (work) is given by

$\Delta U=\int^{t_2}_{t_1}\frac{dU}{dt} dt=\int^{t_2}_{t_1} \frac{d\mathbf{r}}{dt}.m\frac{d^2\mathbf{r}}{dt^2} dt= \int^{\mathbf{r}_2}_{\mathbf{r}_1} m\frac{d^2\mathbf{r}}{dt^2}.d\mathbf{r}=\int^{\mathbf{r}_2}_{\mathbf{r}_1} \mathbf{F}.d\mathbf{r} $

where $\mathbf{F}$ is the force defined as the rate of change of momentum, which for a free particle is $\mathbf{F}=m\frac{d^2\mathbf{r}}{dt^2}$


The relation between energy $E$ and work $W$ is:

$$W=\int dE=\int\frac{dE}{dt}\,dt=\int F\,dx=\int F\,\frac{dx}{dt}\,dt$$

so if you know the enegry you can calculate the work.

for example:

$E=\frac{1}{2} m\,v^2\quad\,\frac{dE}{dt}=m\,v\,\frac{dv}{dt}$

$W=\int m\,v\,\frac{dv}{dt}\,dt=\int m\,a\,\frac{dx}{dt}\,dt=ma\,\int_{x_1}^{x_2}\,dx=ma\,\Delta x$


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