We state the following version of work-energy theorem : $$ K_2-K_1=Fd=W $$ Where acceleration is assumed to be constant, so is the force $F$.

Then the physicists proceed by writing $K_2-K_1=F[x(t_2)-x(t_1)]$ $$=F(x_2-x_1)$$ Notice, $x(t)$ was a polynomial of time $t$ with highest degree of $2$. Now $x_2-x_1$ is just a quantity.

Then they write $K_2-K_1=F \left. x \right|_{x_1}^{x_2}= F \displaystyle \int_{x_1}^{x_2} \, dx $ $$K_2-K_1=\displaystyle \int_{x_1}^{x_2} F \, dx $$

If $G(x)$ is a function such that $\dfrac{\mathrm dG(x)}{\mathrm dx}=F$ then one finds, $$K_2-K_1=G_2-G_1$$ Substituting $G(x)=-U(x)$ yields, $K_2-K_1=-U_2+U_1$ Or this $$E \equiv K_1+U_1=K_2+U_2$$ My question is, isn’t $F$ constant here too? If it is constant then how they apply these equations to non-constant forces like Spring force $F(x)=-kx$?

  • 1
    $\begingroup$ Just do this $\int F\left( x\right) dx$ $\endgroup$
    – Eli
    Jan 28, 2019 at 7:31
  • $\begingroup$ "Where acceleration is assumed to be constant, so is the force F." This assumption is not correct. The principle also applies if you use the average force $F$ applied over the distance $d$. The work-energy principle is used in vehicle crash analysis and seatbelt use. For straight line collisions when you extend the stopping distance you reduce the average impact force on the vehicle and its occupants $\endgroup$
    – Bob D
    Jan 28, 2019 at 13:58

5 Answers 5


Derivation of Work-Energy theorem:

The work done by the force is defined as follows: \begin{equation} W_{12} = \int_{1}^{2} \textbf{F} \cdot d\textbf{s} \end{equation} Assuming the mass is constant, \begin{equation} \int \textbf{F} \cdot d\textbf{s} = m \int \frac{dv}{dt} \cdot \textbf{v} dt = \frac{m}{2} \int \frac{d}{dt} (v^2) dt \end{equation} and therefore,

\begin{equation} W_{12} = \frac{m}{2} (v_{2}^{2} - v_{1}^{2}) = T_2 - T_1 \end{equation} This implies that whether the force is a constant or a variable, when there is a work done on the body, it involves the change in the kinetic energy of the body. Further assuming that the force is conservative, it can be written as gradient of a scalar function. \begin{equation} \textbf{F} = -\nabla \textbf{V} \end{equation} Substituting this is the equation (2), \begin{equation} W_{12} = V_1 - V_2 \end{equation} This implies that, \begin{equation} V_1 + T_1 = V_2 + T_2 \end{equation}


The crucial point lies in the equation itself

Then they write $K_2-K_1=F \left. x \right|_{x_1}^{x_2}= F \displaystyle \int_{x_1}^{x_2} \, dx $

We can take out of the $F$ from the integral only and only if $F$ is not a function of $x$. For example, $F=10N$ is a constant force. It does not depend on time or the position of the particle.

Only in this condition, we can write the above equation. Also, you wrote that

Where acceleration is assumed to be constant ...

So this is the case.

However the equation that you write down ($F=-kx$) depends on $x$ or the position of the particle, so it's not a constant force as you said. In this sense, we cannot write the above equation. Since we cannot take something from out of integral if it depends on the differential part of it. In math,

$$F(x)\int dx\neq \int F(x)dx\,\,(Eqn.1)$$ So, in general, we define the work done as $$W=\int F(x)dx$$

To answer your question in the case of $F=-kx$ clearly a non-constant equation. So we cannot use the derivation that you wrote above but we have to use the general form of the work (Eqn.1)


The force is constant in the result because you start out by assuming it constant from the very beginning. The expression $$Fd=W \qquad \text{(or }\; F(x_2-x_1)=W\text{)}$$ only holds true for constant force. The general expression is $$\int_{x_1}^{x_2} F \;\mathrm dx=W$$ If you start out with that and redo the whole derivation, you'll end up with the same result (and a few of your steps will be redundant) but with no assumption of constant force from the beginning.

  • $\begingroup$ Can you give me a rigorous proof of $W=\displaystyle\int_{x_1}^{x_2} F(x) \, dx$? $\endgroup$
    – user221153
    Jan 28, 2019 at 9:25
  • $\begingroup$ @user221153 Not really a proof, rather a definition. Work is defined as force times the displacement it causes, $F\Delta x$, as you started out with. Now mathematically, we can only do this if the force is constant. If it changes then we have to split the path into smaller segments so that each has a constant force. Then we can multiply force-with-segment-displacement for each segment and add them all up: $$W=\sum F \Delta x$$Those segments may be infinitely small and infinitely many, so we call them $dx$ instead of $\Delta x$ and write it as an integral instead: $$W=\int F\; \mathrm dx$$ $\endgroup$
    – Steeven
    Jan 28, 2019 at 11:31

The definition of work done in moving from position $A$ to position $B$ is $W_{\text{A to B}} =\displaystyle \int_{\rm A}^{\rm B} \vec F(\rm position) \cdot d\vec s$ where $\vec F(\rm position)$ can depend on position ie it does not have to be constant and $d\vec s$ is an incremental displacement of the force.

For a point mass $m$ Newton's second law can be stated as $\vec F =m\vec a = m \dfrac{d\vec v}{dt}$

The work done on the point mass in moving from position $A$ to position $B$ is

$\displaystyle \int_{\rm A}^{\rm B} m \dfrac{d\vec v}{dt} \cdot d\vec s = \int_{\rm A}^{\rm B} m \,d\vec v \cdot \dfrac{d\vec s}{dt} = \int_{\rm A}^{\rm B} m \vec v \cdot d\vec v=\int_{\rm A}^{\rm B} m \left (v_{\rm x} \hat x+v_{\rm y} \hat y+v_{\rm z} \hat z \right)\cdot \left ( dv_{\rm x} \hat x+dv_{\rm y} \hat y+dv_{\rm z} \hat z\right )$

On doing the integration the result is $\frac 12 m v^2_{\rm B} - \frac 12 m v^2_{\rm A}$ which is the change in kinetic energy.

where $v^2_{\rm B} = v^2_{\rm B,x}+v^2_{\rm B,y}+v^2_{\rm B,z}$ and $v^2_{\rm A} = v^2_{\rm A,x}+v^2_{\rm A,y}+v^2_{\rm A,z}$.

So the work done on a point mass is equal to the change in kinetic energy of the mass even if the force is not constant.

Your introduction of potential energy means that you cannot just be considering a point mass and there must be another mass involved, the two masses being the system under consideration.
It is often the case that an assumption is made that the other mass is very much more massive than the point mass $m$ under consideration and so the other mass does not move when the point mass moves.

In such a case the change in the potential energy of the point mass (and the other mass) when mass $m$ moves from position $A$ to position $B$ is $U_{\rm B}-U_{\rm A} =\displaystyle -\int_{\rm A}^{\rm B} \vec F(\rm position) \cdot d\vec s$


Change in kinetic energy is the work done on the body (let us say by a force as a function of distance).

$K_2-K_1 = \int F(x)dx$ $=W$

And $\Delta U = - W$. Which implies $K_1+U_1=K_2+U_2=E$ where $E$ is the total energy.

If your concern is as to how $W$ is derived, then you must realise that it is a definition.


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