# Work done relation to potential energy

I know work done is negative of change in potential energy, I.e., $W=-(∆U)$.

It means that Work done against a force (or work done on a system) increases its potential energy. And Work done by a force (or work done by the system) decreases its potential energy.

But why this is so that an internal force (by internal force I mean that a force created under a system) will always tend to decrease the potential energy of the system while the external force increase the potential energy of the system?

• This formula is only valid in specific cases, where the kinetic energy is zero. Rather use the more general formula of the mechanical energy balance: $$K_1+U_1+W=K_2+U_2$$ – Steeven Jun 12 '16 at 7:24
• Can you give an example of what you mean by your final sentence? – sammy gerbil Jun 12 '16 at 12:53
• @sammygerbil Let's take an example of system of 2 point charges.One be the source charge q and another one be the test charge q' which is +ve. The charge q' will displace towards external force. If source charge is +ve it will repel the q' charge.I have to do +ve work to bring q' from infinity(where PE=0) to finite separation which will increase the PE. If source charge is -ve it will attract the q' charge.I have to do +ve work to displace q' to infinity.So q will do -ve same work t0 bring q' from infinity(where PE=0).-ve work will lower down the PE. – Perspicacious Jun 15 '16 at 9:56
• So Internal force decreases the PE of system whereas External Force increase the PE of system – Perspicacious Jun 15 '16 at 9:57

It is important to note, that you are totally skipping the kinetic Energy $T$ part. The total energy of a system is given by $E=T+U$. There are several examples of systems which build up potential Energy over time. Some examples are:

• The mass of a pendulum is constantly cycling the total energy between kinetic and potential energy

• Objects orbiting a center of mass are generally moving on elliptic orbits. Which are also cycling between potential and kinetic energy.

• A collision of 2 comets can give one of them enough speed to leave the solar system. Thus building up potential energy for eternity.

But your observation does have a valid point. Physical systems tend to go towards lower energy states on their own. So many states which have high potential energy (e.g. a ball on a hill) are unstable. There are several additional questions which cover this topic:

• Is this answering the question? – Steeven Jun 12 '16 at 7:26
• @Steeven I see your point, on a second thought I vote to flag this question as duplicate – user_na Jun 12 '16 at 8:05

The meaning of $$W=-\Delta U$$ has been misinterpreted here. The OP states it as

It means that Work done against a force (or work done on a system) increases its potential energy. And Work done by a force (or work done by the system) decreases its potential energy.

This is false. What this equation means is that the work done by a conservative force is equal to the negative change in potential energy associated with that conservative force. This can easily be seen by using the definition of potential energy and work. Considering a conservative force $$\mathbf F$$ with associated potential energy $$U$$: $$\mathbf F=-\nabla U$$ The work done by this force along some path starting at position $$\mathbf{r_1}$$ and ending at postion $$\mathbf{r_2}$$ is then given by the line integral along this path $$W=\int_{\mathbf{r_1}\rightarrow\mathbf{r_2}}\mathbf F\cdot\text d\mathbf x=-\int_{\mathbf{r_1}\rightarrow\mathbf{r_2}}\nabla U\cdot\text d\mathbf x$$ Using the fundamental theorem of calculus we arrive at $$W=-\left[U(\mathbf{r_2})-U(\mathbf{r_1})\right]=-\Delta U$$

Therefore, this equation is only concerned with thinking about the work done by conservative forces and how that effects the potential energy associated with those forces.