Expanding the concept to a longer path involves summing up the energy required for each path element:

$\bar{W}=\displaystyle\sum_i \vec{F}_i\cdot\Delta\vec{s}_i$

However, the value of this equation represents just an average energy required or generated. The precise energy is obtained when the steps become very small, allowing the force to be considered constant within them:

$W=\displaystyle\sum_i \mbox{lim}_{\Delta\vec{s}_i\rightarrow\vec{0}}\vec{F}_i\cdot\Delta\vec{s}_i$

In this limit, the energy corresponds to the integral along the traveled path, giving us:

En el caso elástico (resorte) la fuerza es



con k la constante del resorte y x la elongación/compresión del resorte. La variación de la energía potencial es

\\n\\nLa diferencia\\n\\n

$\Delta x = x_2 - x_1$

\\n\\ncorresponde al camino recorrido por lo que\\n\\n

$\Delta W=k,x,\Delta x=k(x_2-x_1)\displaystyle\frac{(x_1+x_2)}{2}=\displaystyle\frac{k}{2}(x_2^2-x_1^2)$

y con ello la energía potencial elástica es

The energy required for an object to transition from velocity $v_1$ to velocity $v_2$ can be calculated using the definition with

Using the second law of Newton, this expression can be rewritten as

$\Delta W = m a \Delta s = m\displaystyle\frac{\Delta v}{\Delta t}\Delta s$

Employing the definition of velocity with

we obtain

$\Delta W = m\displaystyle\frac{\Delta v}{\Delta t}\Delta s = m v \Delta v$

where the difference in velocities is

$\Delta v = v_2 - v_1$

Furthermore, the velocity itself can be approximated by the average velocity

$v = \displaystyle\frac{v_1 + v_2}{2}$

Using both expressions, we arrive at

$\Delta W = m v \Delta v = m(v_2 - v_1)\displaystyle\frac{(v_1 + v_2)}{2} = \displaystyle\frac{m}{2}(v_2^2 - v_1^2)$

Thus, the change in energy is given by

$\Delta W = \displaystyle\frac{m}{2}v_2^2 - \displaystyle\frac{m}{2}v_1^2$

In this way, we can define kinetic energy

The energy required for an object to change its angular velocity from $\omega_1$ to $\omega_2$ can be calculated using the definition

Applying Newton's second law, this expression can be rewritten as

$\Delta W=I \alpha \Delta\theta=I\displaystyle\frac{\Delta\omega}{\Delta t}\Delta\theta$

Using the definition of angular velocity

we get

$\Delta W=I\displaystyle\frac{\Delta\omega}{\Delta t}\Delta\theta=I \omega \Delta\omega$

The difference in angular velocities is

$\Delta\omega=\omega_2-\omega_1$

On the other hand, angular velocity itself can be approximated with the average angular velocity

$\omega=\displaystyle\frac{\omega_1+\omega_2}{2}$

Using both expressions, we obtain the equation

$\Delta W=I \omega \Delta \omega=I(\omega_2-\omega_1)\displaystyle\frac{(\omega_1+\omega_2)}{2}=\displaystyle\frac{I}{2}(\omega_2^2-\omega_1^2)$

Thus, the change in energy is given by

$\Delta W=\displaystyle\frac{I}{2}\omega_2^2-\displaystyle\frac{I}{2}\omega_1^2$

This allows us to define kinetic energy as

As the gravitational force is

with $m$ representing the mass. To move this mass from a height $h_1$ to a height $h_2$, a distance of

is covered. Therefore, the energy

with $\Delta s=\Delta h$ gives us the variation in potential energy:

$\Delta W = F\Delta s=mg\Delta h=mg(h_2-h_1)=U_2-U_1=\Delta V$

thus, the gravitational potential energy is