8.1 Potential energy of a system  (Page 2/9)

Basic properties of potential energy

A particle moves along the x -axis under the action of a force given by $F=\text{−}a{x}^2$ , where $a=3{\text{N/m}}^2$ . (a) What is the difference in its potential energy as it moves from $x_A=1\text{m}$ to $x_B=2\text{m}$ ? (b) What is the particle’s potential energy at $x=1\text{m}$ with respect to a given 0.5 J of potential energy at $x=0$ ?

Strategy

(a) The difference in potential energy is the negative of the work done, as defined by [link] . The work is defined in the previous chapter as the dot product of the force with the distance. Since the particle is moving forward in the x -direction, the dot product simplifies to a multiplication ( $\widehat{i}·\widehat{i}=1$ ). To find the total work done, we need to integrate the function between the given limits. After integration, we can state the work or the potential energy. (b) The potential energy function, with respect to zero at $x=0$ , is the indefinite integral encountered in part (a), with the constant of integration determined from [link] . Then, we substitute the x -value into the function of potential energy to calculate the potential energy at $x=1\text{m}.$

Solution

1. The work done by the given force as the particle moves from coordinate x to $x+dx$ in one dimension is
   
   $dW=\overrightarrow{F}·d\overrightarrow{r}=Fdx=\text{−}a{x}^{2}dx.$
   
   Substituting this expression into [link] , we obtain
   
   $\text{Δ}U=\text{−}W={\displaystyle \underset{x_1}{\overset{x_2}{\int }}a{x}^{2}dx}=\frac{1}{3}(3{\text{N/m}}^2){x^2|}_{1\text{m}}^{2\text{m}}=7\text{J}.$
2. The indefinite integral for the potential energy function in part (a) is
   
   $U\left(x\right)=\frac{1}{3}a{x}^3+\text{const}.,$
   
   and we want the constant to be determined by
   
   $U\left(0\right)=0.5\text{J}.$
   
   Thus, the potential energy with respect to zero at $x=0$ is just
   
   $U\left(x\right)=\frac{1}{3}a{x}^3+0.5\text{J}.$
   
   Therefore, the potential energy at $x=1\text{m}$ is
   
   $U\left(1\text{m}\right)=\frac{1}{3}\left(3{\text{N/m}}^2\right){\left(1\text{m}\right)}^3+0.5\text{J}=1.5\text{J}.$

Significance

In this one-dimensional example, any function we can integrate, independent of path, is conservative. Notice how we applied the definition of potential energy difference to determine the potential energy function with respect to zero at a chosen point. Also notice that the potential energy, as determined in part (b), at $x=1\text{m}$ is $U\left(1\text{m}\right)=1\text{J}$ and at $x=2\text{m}$ is $U\left(2\text{m}\right)=8\text{J}$ ; their difference is the result in part (a).

Got questions? Get instant answers now!