Potential Energy of a System

We know that the weight of a body is the downward force of gravity. A ball is released from a particular height in Figure 1. The ball falls under the action of gravitational force of magnitude $mg$ where $m$ is the mass of the ball. The work done by the gravitational force between the heights ${y_1}$ and ${y_2}$ is,

\[\begin{align*} W &= -mg({y_2} - {y_1})\\ {\rm{or,}}\quad W &= mg{y_1} - mg{y_2} = U_1 - U_2 = -(U_2 - U_1) = -\Delta U \tag{1} \label{1} \end{align*}\]

where $U_1$ and $U_2$ are the gravitational potential energies at height $y_1$ and $y_2$ respectively. Therefore, the gravitational potential energy $U$ at any height $y$ is $mgy$ which is the work done by the gravitational force from height $y$ to $y=0$. Notice that the initial potential energy $ U_1 = mg{y_1}$ is the total work done by the gravitational force from $y = {y_1}$ to $y = 0$ and this is similar for the final potential energy $U_2 = mg{y_2}$. So the potential energy is in fact the work done.

Figure 1 A ball falls under the action of downward force of gravity.

Note that $\Delta U$ is the final potential energy $U_2$ minus the initial potential energy $U_1$ but the work done by the gravitational force is the initial potential energy minus the final potential energy. The ball in Figure 1 falls under the action of gravity and therefore $y_1 > y_2$ and the work done by the gravitational force is positive as it should be. The Equation \eqref{1} tells us that the work done by the gravitational force is the negative of change in potential energy.

Mechanics
Physics Key uses third party cookies, read the Terms and Privacy Policy.