Chapter 7, Problem 50P

(a)

To determine

The potential energy function $U\left(x\right)$ associated with force for the system.

Answer to Problem 50P

The potential energy function $U\left(x\right)$ associated with force for the system is $\frac{A{x}^2}{2}-\frac{B{x}^3}{3}$.

Explanation of Solution

Work done by internal forces of a system is equal to negative of change in potential energy of the system.

Write the expression for the conservation of energy.

    $W_{\text{int}}+\Delta U=0$

Here, $W_{\mathrm{int}}$ is the internal energy and $\Delta U$ is the change in potential energy.

Simply the above equation for potential energy.

    $\Delta U=-W_{\mathrm{int}}$                                                                                                                (I)

Write the expression for the change in potential energy.

    $\Delta U=U_f-U_i$

Here, $U_i$ is the initial potential energy and $U_f$  is the final potential energy.

Write the expression for the internal energy.

    $W_{\mathrm{int}}={\displaystyle \underset{x_i}{\overset{x_f}{\int }}Fdx}$

Here, $F$ is the force, $dx$ is the displacement by the force, $x_i$ is the initial position and $x_f$ is the final position.

Substitute ${\displaystyle \underset{x_i}{\overset{x_f}{\int }}Fdx}$ for $W_{\mathrm{int}}$ and $0$ for $U_i$ in equation (I).

    $U_f=-{\displaystyle \underset{x_i}{\overset{x_f}{\int }}{F}_{x}dx}$                                                                                                           (II)

Conclusion:

Substitute $\left(-Ax+B{x}^2\right)\text{N}$ for $F_x$, $0\text{m}$ for $x_i$ and $x$ for $x_f$ in equation (II).

    $\begin{array}{c}{U}_f=-{\displaystyle \underset{0}{\overset{x}{\int }}\left(-Ax+B{x}^2\right)\text{N}dx}\\ ={\displaystyle \underset{0}{\overset{x}{\int }}Ax}dx-{\displaystyle \underset{0}{\overset{x}{\int }}B{x}^{2}dx}\\ =A{\left[\frac{x^2}{2}\right]}_0^x-B{\left[\frac{x^3}{3}\right]}_0^x\\ =\frac{A{x}^2}{2}-\frac{B{x}^3}{3}\end{array}$

Thus, the potential energy function $U\left(x\right)$ associated with force for the system is $\frac{A{x}^2}{2}-\frac{B{x}^3}{3}$.

(b)

To determine

The change in potential energyas the particle moves from $x=2.0\text{m}$ to $x=3.0\text{m}$ .

Answer to Problem 50P

The change in potential energyas the particle moves from $x=2.0\text{m}$ to $x=3.0\text{m}$ is $2.5A-6.33B$.

Explanation of Solution

Write theinitial potential energy expression.

    $U_i\left(x\right)=\frac{A{x}_i^2}{2}-\frac{B{x}_i^3}{3}$.                                                                                                (III)

Here, $U_i\left(x\right)$ is the initial potential energy.

Write thefinal potential energy expression.

    $U_f\left(x\right)=\frac{A{x}_f^2}{2}-\frac{B{x}_f^3}{3}$.                                                                                              (IV)

Here, $U_f\left(x\right)$ is the final potential energy

Write the expression for the change in potential energy.

    $\Delta U=U_f\left(x\right)-U_i\left(x\right)$

Here, $\Delta U$ is the change in potential energy.

Substitute $\frac{A{x}_f^2}{2}-\frac{B{x}_f^3}{3}$ for $U_f\left(x\right)$ and $\frac{A{x}_i^2}{2}-\frac{B{x}_i^3}{3}$ for  $U_i\left(x\right)$ in above equation.

    $\Delta U=\left(\frac{A{x}_f^2}{2}-\frac{B{x}_f^3}{3}\right)-\left(\frac{A{x}_i^2}{2}-\frac{B{x}_i^3}{3}\right)$                                                                         (V)

.

Conclusion:

Substitute $2.0\text{m}$ for $x_i$ and $3.0\text{m}$ for $x_f$ in equation (V).

    $\begin{array}{c}\Delta U=\left(\frac{A{\left(3\right)}^2}{2}-\frac{B{\left(3\right)}^3}{3}\right)-\left(\frac{A{\left(2\right)}^2}{2}-\frac{B{\left(2\right)}^3}{3}\right)\\ =\left(4.5A-9B\right)-\left(2A-2.67B\right)\\ =2.5A-6.33B\end{array}$

Thus, the change in potential energyas the particle moves from $x=2.0\text{m}$ to $x=3.0\text{m}$ is $2.5A-6.33B$.

(c)

To determine

Thechange in kinetic energy as the particle moves from $x=2.0\text{m}$ to $x=3.0\text{m}$.

Answer to Problem 50P

The change in kinetic energy as the particle moves from $x=2.0\text{m}$ to $x=3.0\text{m}$.is $-2.5A+6.33B$.

Explanation of Solution

From the work energy theorem the change in kinetic energy is the work done by the force on the particle.

    $W=\Delta K$                                                                                                                  (VI)

Here $W$ done by particle and $\Delta K$ is change in kinetic energy of the particle.

Substitute $\Delta K$ for $W_{\mathrm{int}}$ in equation (I).

    $\Delta U=-\Delta K$

Simplify the above equation for change in kinetic energy.

    $\Delta K=-\Delta U$                                                                                                            (VII)

Conclusion:

Substitute $2.5A-6.33B$ for $\Delta U$ in equation (VII).

    $\begin{array}{c}\Delta K=-\left(2.5A-6.33B\right)\\ =-2.5A+6.33B\end{array}$

Thus, the change in kinetic energy as the particle moves from $x=2.0\text{m}$ to $x=3.0\text{m}$. is $-2.5A+6.33B$.