Chapter 7, Problem 22P

(a)

To determine

The potential energy function.

Explanation of Solution

Given:

The value of constant force is $6.0\text{N}$ .

The initial potential energy is $0\text{J}$ .

Formula used:

  $U\left(x\right)$ is the potential energy of the particle and $F_x$ is negative of derivative of potential energy function.

Write the expression for force acting on the object.

  $F_x=-\frac{dU}{dx}$

Here, $F_x$ is the force acting on the object, $dU$ is change of potential energy and $dx$ is change in position.

Rearrange the above expression in terms of $dU$ and integrate it.

  ${\displaystyle \underset{x_0}{\overset{x}{\int }}dU}=-{\displaystyle \underset{x_0}{\overset{x}{\int }}{F}_{x}dx}$

Simplify the above expression in terms of $U\left(x\right)$ .

  $U\left(x\right)=U\left(x_0\right)-F_x\left(x-x_0\right)$ .......... (1)

Here, $U\left(x\right)$ is the final energy, $U\left(x_0\right)$ is the initial energy, $x$ is the final position and $x_0$ is the initial position.

Calculation:

Substitute $0\text{J}$ for $U\left(x_0\right)$ and $6.0\text{N}$ for $F_x$ in equation (I).

  $\begin{array}{c}U\left(x\right)=\left(0\text{J}\right)-\left(6.0\text{N}\right)\left(x-x_0\right)\\ =-\left(6.0\text{N}\right)\left(x-x_0\right)\end{array}$

Conclusion:

Thus, the potential energy function is $U\left(x\right)=-\left(6.0\text{N}\right)\left(x-x_0\right)$ .

(b)

To determine

The potential energy function.

Explanation of Solution

Given:

The value of constant force is $6.0\text{N}$ .

The initial potential energy is $0\text{J}$ .

The potential energy at $x=4.0\text{m}$ is $0\text{J}$ .

Formula used:

  $U\left(x\right)$ is the potential energy of the particle and $F_x$ is negative of derivative of potential energy function.

Write the expression for force acting on the object.

  $F_x=-\frac{dU}{dx}$

Here, $F_x$ is the force acting on the object, $dU$ is change of potential energy and $dx$ is change in position.

Rearrange the above expression in terms of $dU$ and integrate it.

  ${\displaystyle \underset{x_0}{\overset{x}{\int }}dU}=-{\displaystyle \underset{x_0}{\overset{x}{\int }}{F}_{x}dx}$

Simplify the above expression in terms of $U\left(x\right)$ .

  $U\left(x\right)=U\left(x_0\right)-F_x\left(x-x_0\right)$ .......... (1)

Here, $U\left(x\right)$ is the final energy, $U\left(x_0\right)$ is the initial energy, $x$ is the final position and $x_0$ is the initial position.

Calculation:

Substitute $0\text{J}$ for $U\left(x_0\right)$ , $0\text{J}$ for $U\left(x\right)$ , $4.0\text{m}$ for $x$ and $6.0\text{N}$ for $F_x$ in equation (I).

  $\begin{array}{c}0\text{J}=\left(0\text{J}\right)-\left(6.0\text{N}\right)\left(4.0\text{m}-x_0\right)\\ 0=4.0\text{m}-x_0\\ x_0=4.0\text{m}\end{array}$

Substitute $0\text{J}$ for $U\left(x_0\right)$ , $4.0\text{m}$ for $x_0$ and $6.0\text{N}$ for $F_x$ in equation (I).

  $\begin{array}{c}U\left(x\right)=\left(0\text{J}\right)-\left(6.0\text{N}\right)\left(x-4.0\text{m}\right)\\ =-\left(6.0\text{N}\right)\left(x-4.0\text{m}\right)\\ =24\text{J}-\left(6.0\text{N}\right)x\end{array}$

Conclusion:

Thus, the potential energy function is $U\left(x\right)=24\text{J}-\left(6.0\text{N}\right)x$ .

(c)

To determine

The potential energy function.

Explanation of Solution

Given:

The value of constant force is $6.0\text{N}$ .

The initial potential energy is $0\text{J}$ .

The potential energy at $x=6.0\text{m}$ is $14\text{J}$ .

Formula used:

  $U\left(x\right)$ is the potential energy of the particle and $F_x$ is negative of derivative of potential energy function.

Write the expression for force acting on the object.

  $F_x=-\frac{dU}{dx}$

Here, $F_x$ is the force acting on the object, $dU$ is change of potential energy and $dx$ is change in position.

Rearrange the above expression in terms of $dU$ and integrate it.

  ${\displaystyle \underset{x_0}{\overset{x}{\int }}dU}=-{\displaystyle \underset{x_0}{\overset{x}{\int }}{F}_{x}dx}$

Simplify the above expression in terms of $U\left(x\right)$ .

  $U\left(x\right)=U\left(x_0\right)-F_x\left(x-x_0\right)$ .......... (1)

Here, $U\left(x\right)$ is the final energy, $U\left(x_0\right)$ is the initial energy, $x$ is the final position and $x_0$ is the initial position.

Calculation:

Substitute $0\text{J}$ for $U\left(x_0\right)$ , $14\text{J}$ for $U\left(x\right)$ , $6.0\text{m}$ for $x$ and $6.0\text{N}$ for $F_x$ in equation (I).

  $\begin{array}{c}14\text{J}=\left(0\text{J}\right)-\left(6.0\text{N}\right)\left(6.0\text{m}-x_0\right)\\ 14\text{J}=-36\text{J+}\left(6.0\text{N}\right)x_0\\ x_0=\frac{25}{3}\text{m}\end{array}$

Substitute $0\text{J}$ for $U\left(x_0\right)$ , $\frac{25}{3.0}\text{m}$ for $x_0$ and $6.0\text{N}$ for $F_x$ in equation (I).

  $\begin{array}{c}U\left(x\right)=\left(0\text{J}\right)-\left(6.0\text{N}\right)\left(x-\frac{25}{3.0}\text{m}\right)\\ =50\text{J}-\left(6.0\text{N}\right)x\end{array}$

Conclusion:

Thus, the potential energy function is $U\left(x\right)=50\text{J}-\left(6.0\text{N}\right)x$ .