Examples of Energy Transfer in an Isolated System

An isolated system is one where energy nor matter is transferred into or out of the system.

Earth & Ball §

Take the isolated system of a ball above the Earth’s surface, and the Earth. In this system:

• This system has no external forces, and as is considered an isolated system, which means it has energy conserved i.e. no change in total energy between configurations. In other words, $\Delta E_{sys}=0$
• Potential energy comes into play, because we have 2 objects in a system, both with mass. As such, the fundamental force of gravity is inside the system, which results in potential energy in the system.

In terms of Energy §

Recall that the total energy in a system is:

$$E_{sys}=\sum K_{objects}+U_{sys}$$

In configuration 1, there is no kinetic energy because no objects have velocity. So:

$$K_{sys}=\sum K_{objects}\to K_i=0$$

Gravitational potential energy is non-zero, because we have a system of two masses, which are a finite distance apart:

$$U_{sys}=U_{grav}$$

So our total energy is:

$$E_{sy{s}_i}=U_i+K_i=U_{gra{v}_i}+0$$

In configuration 2, there is kinetic energy in the ball, and almost zero kinetic energy in the Earth (because it barely moves). In order to satisfy energy conservation, this kinetic energy is converted from the potential energy. So:

$$E_{sy{s}_f}=K_{ball}+K_{earth}+U_{gra{v}_f}$$

Recall that the change in energy, $\Delta E=0$. So the final and initial energy of the system must be the same:

$$E_{sy{s}_i}=E_{sy{s}_f}$$

$$U_{gra{v}_i}+K_i=K_{ball}+K_{earth}+U_{gra{v}_f}$$

Finally, with some rearranging:

$$U_{gra{v}_i}-U_{gra{v}_f}=K_f-K_i$$

$$-\Delta U=\Delta K$$

And:

$$\Delta K+\Delta U=0=\Delta E$$

In terms of Forces and Work Done §

In configuration 1 and 2, there exists a constant gravitational force on the ball by the Earth, and also an equal but opposite force on the Earth by the ball.

When the ball falls, positive work is done on it, as the force of gravity is in the same direction as that of the displacement, and $W=\vec{F}\cdot \vec{x}$. As such, $W>0$.

Through the Work-Energy Theorem, positive work implies a positive change in Kinetic energy:

$$W=\Delta K$$

However, because this work done was a result of a conservative force, it also results in the negative change in Potential Energy:

$$W=-\Delta U$$

Which means work is done within the system while still obeying the laws of conservation, since the increase in kinetic energy directly corresponds with a decrease in potential energy.