Chapter 7, Problem 44P

(a)

To determine

The force is conservative.

Answer to Problem 44P

Yes, the force is conservative.

Explanation of Solution

Conservative forces are the forces whose work done depends on the initial and final position and not on the path followed.

When the particle is pushed by a variable force, work is done on it by the applied force to move it a certain distance ahead.

Write the expression for work done by a variable force to displace a particle.

  $W_{ext}={\displaystyle \underset{r_i}{\overset{r_f}{\int }}\overrightarrow{F}\cdot d\overrightarrow{r}}$                                                                                                            (I)

Here, $W_{ext}$ is the work done, $F$ is the external force applied;$\overrightarrow{r}$ is the displacement vector of the particle, $r_i$ is the initial position and $r_f$ is the final position.

Consider force to be constant and solve the above equation (I).

    $\begin{array}{c}{W}_{ext}=\overrightarrow{F}\cdot {\displaystyle \underset{r_i}{\overset{r_f}{\int }}d\overrightarrow{r}}\\ =\overrightarrow{F}\cdot \left({\overrightarrow{r}}_f-{\overrightarrow{r}}_i\right)\end{array}$

Therefore, the work done is a function of end states only and not the path followed.

Conclusion:

Thus, the force is conservative.

(b)

To determine

The work done by $\overrightarrow{F}$ on the particle as it moves along each one of the three paths.

Answer to Problem 44P

The work done by $\overrightarrow{F}$ on the particle as it moves along all of the three paths is $35\text{J}$.

Explanation of Solution

Consider three paths across which force acts. The three paths are $OAC$, $OC$ and $OBC$.

Write the expression for the work done along path $OAC$.

    $W_1={\displaystyle \underset{O}{\overset{A}{\int }}\overrightarrow{F}d\overrightarrow{r}}+{\displaystyle \underset{A}{\overset{C}{\int }}\overrightarrow{F}d\overrightarrow{r}}$                                                                                              (II)

Here, $W_1$ is the work done along path $OAC$, $OA$ is the length in x direction and $AC$ is the length in y direction.

Write the expression for the work done along path $OC$.

    $W_2={\displaystyle \underset{O}{\overset{C}{\int }}\overrightarrow{F}d\overrightarrow{r}}$                                                                                                          (III)

Here, $W_2$ is the work done along path $OC$ and $OC$ is the length in x-y direction.

Write the expression for the work done along path $OBC$.

    $W_3={\displaystyle \underset{O}{\overset{B}{\int }}\overrightarrow{F}d\overrightarrow{r}}+{\displaystyle \underset{B}{\overset{C}{\int }}\overrightarrow{F}d\overrightarrow{r}}$                                                                                            (IV)

Here, $W_3$ is the work done along path $OBC$, $OB$ is the length in y direction and $BC$ is the length in x direction.

The displacement vector $\overrightarrow{r}$ is along x-y axis.

Write the expression for displacement vector.

    $\overrightarrow{r}=x\widehat{i}+y\widehat{j}$

Conclusion:

Substitute $\left(3\widehat{i}+4\widehat{j}\right)\text{N}$ for $\overrightarrow{F}$, $5$ for length of $OA$, $5$ for length of $AC$ and $x\widehat{i}+y\widehat{j}$ for $\overrightarrow{r}$ in equation (II).

    $\begin{array}{c}{W}_1={\displaystyle \underset{O}{\overset{A}{\int }}\left(3\widehat{i}+4\widehat{j}\right)Nd\left(x\widehat{i}+y\widehat{j}\right)}+{\displaystyle \underset{A}{\overset{C}{\int }}\left(3\widehat{i}+4\widehat{j}\right)Nd\left(x\widehat{i}+y\widehat{j}\right)}\\ ={\displaystyle \underset{0}{\overset{5}{\int }}\left(3\widehat{i}dx\widehat{i}+4\widehat{j}dy\widehat{j}\right)}+{\displaystyle \underset{0}{\overset{5}{\int }}\left(3\widehat{i}dx\widehat{i}+4\widehat{j}dy\widehat{j}\right)}\\ =\left(3{\left[x\right]}_0^5+0\right)+\left(0+4{\left[y\right]}_0^5\right)\end{array}$

Simplify the above equation.

    $\begin{array}{c}{W}_1=\left(3{\left[x\right]}_0^5+0\right)+\left(0+4{\left[y\right]}_0^5\right)\\ =3\left(5-0\right)+4\left(5-0\right)\\ =35\text{J}\end{array}$

Substitute $\left(3\widehat{i}+4\widehat{j}\right)N$ for $\overrightarrow{F}$ and $x\widehat{i}+y\widehat{j}$ for $\overrightarrow{r}$  in equation (III).

    $\begin{array}{c}{W}_2={\displaystyle \underset{O}{\overset{C}{\int }}\left(3\widehat{i}+4\widehat{j}\right)Nd\left(x\widehat{i}+y\widehat{j}\right)}\\ ={\displaystyle \underset{0}{\overset{5}{\int }}3\widehat{i}dx\widehat{i}+{\displaystyle \underset{0}{\overset{5}{\int }}4\widehat{j}dy\widehat{j}}}\\ =3{\left[x\right]}_0^5+4{\left[y\right]}_0^5\\ =35\text{J}\end{array}$

Substitute $\left(3\widehat{i}+4\widehat{j}\right)\text{N}$ for $\overrightarrow{F}$, $5$ for length of $OB$, $5$ for length of $BC$ and $x\widehat{i}+y\widehat{j}$ for $\overrightarrow{r}$  in equation (IV).

    $\begin{array}{c}{W}_3={\displaystyle \underset{O}{\overset{B}{\int }}\left(3\widehat{i}+4\widehat{j}\right)Nd\left(x\widehat{i}+y\widehat{j}\right)}+{\displaystyle \underset{B}{\overset{C}{\int }}\left(3\widehat{i}+4\widehat{j}\right)Nd\left(x\widehat{i}+y\widehat{j}\right)}\\ ={\displaystyle \underset{y=0}{\overset{y=5}{\int }}\left(3\widehat{i}dx\widehat{i}+4\widehat{j}dy\widehat{j}\right)}+{\displaystyle \underset{x=0}{\overset{x=5}{\int }}\left(3\widehat{i}dx\widehat{i}+4\widehat{j}dy\widehat{j}\right)}\\ =\left(0+4{\left[y\right]}_0^5\right)+\left(3{\left[x\right]}_0^5+0\right)\end{array}$

Simplify the above equation.

    $\begin{array}{c}{W}_3=\left(0+4{\left[y\right]}_0^5\right)+\left(3{\left[x\right]}_0^5+0\right)\\ =4\left(5-0\right)+3\left(5-0\right)\\ =35\text{J}\end{array}$

The work done by the force on the path $OAC$ is $35\text{J}$.

The work done by the force on the path $OC$ is $35\text{J}$.

The work done by the force on the path $OBC$ is $35\text{J}$.

Thus, the work done by $\overrightarrow{F}$ on the particle as it moves along all of the three paths is $35\text{J}$ respectively.