I am studying the work energy therom and I understand that 1/2(mv^2)=Wnet, however, I saw this picture below online.

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I understand the summation of the force times the change in distance. However, why is the integral there?


Because it's not the force times the total distance, it's the integral of the force with respect to distance, which is like saying it's the sum of the (Force x Distance) values at each infinitesimally small distance. This is because the force may not be constant over the distance. If it was, then it would simply be the (Constant force x distance), and not the integral. If the force was 1N for 4m and 2N for 7m, the work would be (1+1+1+1+2+2+2+2+2+2+2+2) joules, measuring the work in 1 meter bins (i.e. each number is 1m x the force AT THAT METER ALONE). This allows for the work to be computed when the force changes. The integral is analogous to this except the distance bins are infinitesimally small.


It is physicists that defined work as $dW=\vec F \cdot d \vec r$. The reason is well explained by Jory. The force $\vec F$ can change value or direction through motion. So we calculate the work for small (infinitesimal) transpositions that the vector of force is constrant. If we sum all the infinitesimal works then we have the whole work that the force produces. Of course this sum is calculated with the integral.

$W=\int_0^W dW $

$\int \vec F \cdot d\vec r = \int m\vec a \cdot d \vec r = \int m \frac{d \vec v}{dt} \cdot d \vec r=\int md\vec v \cdot \frac{d \vec r}{dt}=\int m\vec v\cdot d \vec v= Δ(\frac{1}{2}mv^2)$

The quantity $T= \frac{1}{2}mv^2$ we call Kinetic Energy. Therefore a theorem is dercribed by the above relations, which we call Work Energy Theorem that states:


Now the integral of motions were introduced to make the problem solving easier. Even the solution of the equations of motion for a system of three particles are quite difficult, let alone more. So physicists found,experimentally, quantities that are functions of the position and velocity of the particles and that are constant through motion if the system is isolated-constants of motion.

One of these quantities is energy and has the additive property, meaning that:

$\sum W= Δ(\frac{1}{2}mv^2)_{total}$

Let's consider a cituation that many forces are acted on a partice. Then each one of them makes an impact on it, it produces work that change the kinetic energy. So the total change of the kinetic energy is equal to the sum of the changes by each force.

$\int \sum \vec F \cdot d \vec r = \int ( \vec F_1 + \vec F_2 +...+ \vec F_i) \cdot d \vec r= \int \vec F_1 \cdot d \vec r + \int \vec F_2 \cdot d \vec r +...+\int \vec F_i \cdot d \vec r$


$Δ(\frac{1}{2}mv^2)_{total}=\sum_i \int \vec F_i \cdot d \vec r$


Because the Work in the theorem is the total work due to all external forces acting on the system. The work due to a single force is simply one integral term (which is what you have inside the sum) and when you need to calculate the total work, then you need to add all such integral terms. Thanks,

  • $\begingroup$ Why is the force calculated by the integral of the force? $\endgroup$ – eli Mar 6 '16 at 6:43
  • $\begingroup$ It is the work due to a single force that is calculated by the integral of that force. So, the integral term you expressed inside the sum is only the work due to a single force as I said before. By definition, the integral of any force is the work due to that force. $\endgroup$ – Benjamin Mar 6 '16 at 6:47
  • $\begingroup$ With Jory's explanation, your question is in fact answered completely. As pointed out, each one of the forces in general is not constant and that is why we need to add the infinitesimal contributions. Thanks to Newton for the introduction of Calculus where by an integration we really mean an infinite summation. But, don't confuse the integral with the summation in your case. The first summation is a sum of a finite number of works. Yet the integration is actually an infinite summation when it comes to mathematics. $\endgroup$ – Benjamin Mar 6 '16 at 7:06

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