Chapter 7, Problem 31P

(a)

To determine

The points where object is in equilibrium.

Explanation of Solution

Given:

The potential energy function is $U=3x^2-x^3$ for $x\le 3.0\text{m}$ .

The potential energy function is $U=0$ for $x\ge 3.0\text{m}$ .

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 in terms of potential energy.

  $F_x=-\frac{dU}{dx}$ ........(1)

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

Calculation:

For $x\le 3.0\text{m}$ :

Substitute $3x^2-x^3$ for $U$ in equation (1).

  $\begin{array}{c}{F}_x=-\frac{d}{dx}\left(3x^2-x^3\right)\\ =3x\left(2-x\right)\end{array}$

Calculate the equilibrium points.

  $\begin{array}{c}{F}_x=0\\ 3x\left(2-x\right)=0\\ \Rightarrow x=0\text{and}x=2.0\text{m}\end{array}$

For $x\ge 3.0\text{m}$ :

Substitute $0$ for $U$ in equation (1).

  $\begin{array}{c}{F}_x=-\frac{d}{dx}\left(0\right)\\ =0\end{array}$

Object would be in neutral equilibrium for $x\ge 3.0\text{m}$ .

Conclusion:

Thus, object is in equilibrium at $x=0,x=2.0\text{m and }x\ge 3.0\text{m}$ .

(b)

To determine

The graph for $U$ versus $x$ .

Explanation of Solution

Given:

The potential energy function is $U=3x^2-x^3$ for $x\le 3.0\text{m}$ .

The potential energy function is $U=0$ for $x\ge 3.0\text{m}$ .

Formula used:

The energy contained in an object due to its position when compared to other objects is called potential energy.

Write the expression of potential energy for $x\le 3.0\text{m}$ .

  $U=3x^2-x^3$ ........(2)

Here, $U$ is potential energy, $x$ is the position of object.

Write the expression of potential energy for $x\ge 3.0\text{m}$ .

  $U=0$

Calculation:

Substitute $-1.0\text{m}$ for $x$ in equation (2).

  $\begin{array}{c}U=\left(3.0\text{J}/{\text{m}}^2\right){\left(-1.0\text{m}\right)}^2-\left(1.0\text{J}/{\text{m}}^2\right){\left(-1.0\text{m}\right)}^3\\ =4.0\text{J}\end{array}$

Substitute $0\text{m}$ for $x$ in equation (2).

  $\begin{array}{c}U=\left(3.0\text{J}/{\text{m}}^2\right){\left(0\text{m}\right)}^2-\left(1.0\text{J}/{\text{m}}^2\right){\left(0\text{m}\right)}^3\\ =0\text{J}\end{array}$

Substitute $1.0\text{m}$ for $x$ in equation (2).

  $\begin{array}{c}U=\left(3.0\text{J}/{\text{m}}^2\right){\left(1.0\text{m}\right)}^2-\left(1.0\text{J}/{\text{m}}^2\right){\left(1.0\text{m}\right)}^3\\ =2.0\text{J}\end{array}$

Substitute $2.0\text{m}$ for $x$ in equation (2).

  $\begin{array}{c}U=\left(3.0\text{J}/{\text{m}}^2\right){\left(2.0\text{m}\right)}^2-\left(1.0\text{J}/{\text{m}}^2\right){\left(2.0\text{m}\right)}^3\\ =4.0\text{J}\end{array}$

Substitute $3.0\text{m}$ for $x$ in equation (2).

  $\begin{array}{c}U=\left(3.0\text{J}/{\text{m}}^2\right){\left(3.0\text{m}\right)}^2-\left(1.0\text{J}/{\text{m}}^2\right){\left(3.0\text{m}\right)}^3\\ =0\text{J}\end{array}$

The graph for $U$ versus $x$ is plotted below.

  

Conclusion:

Thus, the graph for $U$ versus $x$ is plotted above.

(c)

To determine

The stable and unstable points of equilibrium.

Explanation of Solution

Given:

The potential energy function is $U=3x^2-x^3$ for $x\le 3.0\text{m}$ .

The potential energy function is $U=0$ for $x\ge 3.0\text{m}$ .

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 in terms of potential energy.

  $F_x=-\frac{dU}{dx}$ ........(1)

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

Calculation:

For $x\le 3.0\text{m}$ :

Substitute $3x^2-x^3$ for $U$ in equation (1).

  $\begin{array}{c}{F}_x=-\frac{d}{dx}\left(3x^2-x^3\right)\\ =3x\left(2-x\right)\end{array}$

Calculate the equilibrium points.

  $\begin{array}{c}{F}_x=0\\ 3x\left(2-x\right)=0\\ \Rightarrow x=0\text{and}x=2.0\text{m}\end{array}$

For $x\ge 3.0\text{m}$ :

Substitute $0$ for $U$ in equation (1).

  $\begin{array}{c}{F}_x=-\frac{d}{dx}\left(0\right)\\ =0\end{array}$

Object would be in neutral equilibrium for $x\ge 3.0\text{m}$ .

Calculate the potential energy at all equilibrium points.

For $x=0$ :

  $\begin{array}{c}U=\left(3.0\text{J}/{\text{m}}^2\right){\left(0\text{m}\right)}^2-\left(1.0\text{J}/{\text{m}}^2\right){\left(0\text{m}\right)}^3\\ =0\text{J}\end{array}$

For $x=2.0\text{m}$ :

  $\begin{array}{c}U=\left(3.0\text{J}/{\text{m}}^2\right){\left(2.0\text{m}\right)}^2-\left(1.0\text{J}/{\text{m}}^2\right){\left(2.0\text{m}\right)}^3\\ =4.0\text{J}\end{array}$

Conclusion:

The value of potential energy is maximum at $x=2.0\text{m}$ and minimum at $x=0$ . Thus, the stable equilibrium point is $x=0$ and the unstable equilibrium point is $x=2.0\text{m}$.

(d)

To determine

The speed of object at $x=2.0\text{m}$ .

Explanation of Solution

Given:

The potential energy function is $U=3x^2-x^3$ for $x\le 3.0\text{m}$ .

The potential energy function is $U=0$ for $x\ge 3.0\text{m}$ .

The mass of object is $4.0\text{kg}$ .

The total energy of object is $12\text{J}$ at $x=2.0\text{m}$ .

Formula used:

Write the expression for potential energy.

  $U=3x^2-x^3$ ........(2)

Here, $x$ is the position of object and $U$ is the potential energy.

Write the expression for change in kinetic energy.

  $K=\frac{1}{2}m{v}^2$

Here, $m$ is mass of object, $v$ is velocity of object and $K$ is the kinetic energy.

Write the expression of total energy.

  $E=U+K$ ........(3)

Here, $E$ is the total energy.

Substitute $\frac{1}{2}m{v}^2$ for $K$ in equation (3) and rearrange in terms of $v$ .

  $v=\sqrt{\frac{2\left(E-U\right)}{m}}$ ........(4)

Calculation:

Substitute $2.0\text{m}$ for $x$ in equation (2).

  $\begin{array}{c}U=\left(3.0\text{J}/{\text{m}}^2\right){\left(2.0\text{m}\right)}^2-\left(1.0\text{J}/{\text{m}}^2\right){\left(2.0\text{m}\right)}^3\\ =4.0\text{J}\end{array}$

Substitute $12\text{J}$ for $E$ , $4.0\text{J}$ for $U$ and $4.0\text{kg}$ for $m$ in equation (4).

  $\begin{array}{c}v=\sqrt{\frac{2\left(12\text{J}-4.0\text{J}\right)}{4.0\text{kg}}}\\ =2.0\text{m}/\text{s}\end{array}$

Conclusion:

The speed of object at $x=2.0\text{m}$ is $2.0\text{m}/\text{s}$ .