Chapter 7, Problem 32P

(a) Suppose a constant force acts on an object. The force does not vary with time or with the position or the velocity of the object. Start with the general definition for work done by a force

$W={\displaystyle {\int }_i^f\stackrel{\to }{F}\cdot d\stackrel{\to }{r}}$

and show that the force is conservative. (b) As a special case, suppose the force $\stackrel{\to }{F}=(3\hat{i}+4\hat{j})$ N acts on a panicle that moves from O to © in Figure P7.31. Calculate the work done by $\stackrel{\to }{F}$ on the particle as it moves along each one of the three paths shown in the figure and show that the work done along the three paths is identical. (c) What If? Is the work done also identical along the three paths for the force $\stackrel{\to }{F}=(4x\hat{i}+3y\hat{j})$, where $\stackrel{\to }{F}$ is in newtons and x and y are in meters, from Problem 19? (d) What If? Suppose the force is given by $\stackrel{\to }{F}=(y\hat{i}+x\hat{j})$, where $\stackrel{\to }{F}$ is in newtons and x and y are in meters. Is the work done identical along the three paths for this force?

To determine

That the constant force act on the object is conservative.

Answer to Problem 32P

The constant force applied on the object is conservative in nature.

Explanation of Solution

 The work done by a conservative force on a particle moving between any two points is independent of the path taken by the particle. Its only depends upon the end points of the path taken by the particle to move.

The general definition for work done by a force

    $W={\displaystyle {\int }_i^f\overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$.

Write the formula to calculate the work done by the force on the object

    $W={\displaystyle {\int }_i^f\overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$

Here, $W$ is the work done by the force on the particle, $\overrightarrow{\text{F}}$ is the applied force vector on the object and $d\overrightarrow{\text{r}}$ is the displacement vector of the object.

Since the force is constant that does not vary with respect to time or the position or the velocity of the object. So, the value of force can be taken out from the integration since it is constant quantity.

    $\begin{array}{c}W=\overrightarrow{\text{F}}{\displaystyle {\int }_i^{f}d\overrightarrow{\text{r}}}\\ =\overrightarrow{\text{F}}{\left[\overrightarrow{\text{r}}\right]}_i^f\\ =\overrightarrow{\text{F}}\left[{\overrightarrow{\text{r}}}_{\text{f}}-{\overrightarrow{\text{r}}}_{\text{i}}\right]\end{array}$

Now, here the force is constant so, the work done by this force on the object in only depends upon the end points of the displace object that shows the work done is independent of the path taken by the object to displace between the end points. But the work done is independent of the path only when the force is conservative.

Conclusion:

Therefore, the constant force applied on the object is conservative in nature.

To determine

The work done by the force $\overrightarrow{\text{F}}=\left(3\widehat{\text{i}}\text{+4}\widehat{\text{j}}\right)\text{N}$ on the particle along each one of the three paths shown in the figure and show that the work done along the three path is identical.

Answer to Problem 32P

The work done by the force $\overrightarrow{\text{F}}=\left(3\widehat{\text{i}}\text{+4}\widehat{\text{j}}\right)\text{N}$ on the particle along all the three path are $35\text{J}$ and the work done by the force is identical in all the three paths.

Explanation of Solution

The given  force is $\overrightarrow{\text{F}}=\left(3\widehat{\text{i}}\text{+4}\widehat{\text{j}}\right)\text{N}$.

The force acting on the particle is $\overrightarrow{\text{F}}=\left(3\widehat{\text{i}}\text{+4}\widehat{\text{j}}\right)\text{N}$ and the three different path of the particle moves is shown in figure (I).

Figure (I)

Write the formula to calculate the work done by the force on the particle

    $W={\displaystyle \int \overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$

Substitute $\left(3\widehat{\text{i}}\text{+4}\widehat{\text{j}}\right)\text{N}$ for $\overrightarrow{\text{F}}$ and $dx\widehat{\text{i}}$ for $d\widehat{\text{r}}$.

    $\begin{array}{c}W={\displaystyle \int \left(\left(3\widehat{\text{i}}\text{+4}\widehat{\text{j}}\right)\text{N}\right)\cdot \left(dx\widehat{\text{i}}\right)}\text{m}\\ ={\displaystyle \int \left(3dx\right)\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}}\\ =\left[{\displaystyle \int 3dx}\right]\text{J}\end{array}$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00\text{m}$ and there is no displacement in y-direction that is $y=0$.

Taking the limits of the integration,

    $\begin{array}{c}{W}_{\text{OA}}=\left[{\displaystyle {\int }_0^{5.00}3dx}\right]\text{J}\\ =\left(3{\left[x\right]}_0^{5.00}\right)\text{J}\\ =\left(3\left[5.00-0\right]\right)\text{J}\\ =15\text{J}\end{array}$

In the path $AC$, the particle moves from $y=0$ to $y=5.00\text{m}$ and there is a displacement of particle in x-direction that is $x=5.00\text{m}$.

Write the formula to calculate the work done by the force on the particle

    $W={\displaystyle \int \overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$

Substitute $\left(3\widehat{\text{i}}\text{+4}\widehat{\text{j}}\right)\text{N}$ for $\overrightarrow{\text{F}}$ and $dy\widehat{\text{j}}$ for $d\widehat{\text{r}}$.

    $\begin{array}{c}W={\displaystyle \int \left(\left(3\widehat{\text{i}}\text{+4}\widehat{\text{j}}\right)\text{N}\right)\cdot \left(dy\widehat{\text{j}}\right)}\text{m}\\ ={\displaystyle \int \left(4dy\right)\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}}\\ =\left[{\displaystyle \int 4dy}\right]\text{J}\end{array}$

Taking the limits of the integration,

    $\begin{array}{c}{W}_{\text{AC}}=\left[{\displaystyle {\int }_0^{5.00}4dy}\right]\text{J}\\ \text{=}\left(\text{4}{\left[y\right]}_0^{5.00}\right)\text{J}\\ =\text{4}\left[5.00-0\right]\\ =20\text{J}\end{array}$

Write the formula to calculate the work done by the force on the particle along the purple path $OAC$

    $W_{\text{purple}}=W_{\text{OA}}+W_{\text{AC}}$

Here, $W_{\text{purple}}$ is the work done along the purple path, $W_{\text{OA}}$ is the work done along the path $OA$ and $W_{\text{AC}}$ is the work done along the path $AC$.

Substitute $15\text{J}$ for $W_{\text{OA}}$ and $20\text{J}$ for $W_{\text{AC}}$.

    $\begin{array}{c}{W}_{\text{purple}}=15\text{J}+20\text{J}\\ =35\text{J}\end{array}$

Write the formula to calculate the work done by the force on the particle

    $W={\displaystyle \int \overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$

Substitute $\left(3\widehat{\text{i}}\text{+4}\widehat{\text{j}}\right)\text{N}$ for $\overrightarrow{\text{F}}$ and $dy\widehat{\text{j}}$ for $d\widehat{\text{r}}$.

    $\begin{array}{c}W={\displaystyle \int \left(\left(3\widehat{\text{i}}\text{+4}\widehat{\text{j}}\right)\text{N}\right)\cdot \left(dy\widehat{\text{j}}\right)}\text{m}\\ ={\displaystyle \int \left(4dy\right)\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}}\\ =\left[{\displaystyle \int 4dy}\right]\text{J}\end{array}$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00\text{m}$ and there is no displacement in x-direction that is $x=0$.

Taking the limits of the integration,

    $\begin{array}{c}{W}_{\text{OB}}=\left[{\displaystyle {\int }_0^{5.00}4dy}\right]\text{J}\\ =4{\left[y\right]}_0^{5.00}\\ =\left(4\left[5.00-0\right]\right)\text{J}\\ =20\text{J}\end{array}$

Write the formula to calculate the work done by the force on the particle

    $W={\displaystyle \int \overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$

Substitute $\left(3\widehat{\text{i}}\text{+4}\widehat{\text{j}}\right)\text{N}$ for $\overrightarrow{\text{F}}$ and $dx\widehat{\text{i}}$ for $d\widehat{\text{r}}$.

    $\begin{array}{c}W={\displaystyle \int \left(\left(3\widehat{\text{i}}\text{+4}\widehat{\text{j}}\right)\text{N}\right)\cdot \left(dx\widehat{\text{i}}\right)}\text{m}\\ ={\displaystyle \int \left(3dx\right)\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}}\\ =\left[{\displaystyle \int 3dx}\right]\text{J}\end{array}$

In the path $BC$, the particle moves from $x=0$ to $x=5.00\text{m}$ and there is a displacement of particle in y-direction that is $y=5.00\text{m}$.

Taking the limits of the integration,

    $\begin{array}{l}{W}_{\text{BC}}=\left[{\displaystyle {\int }_0^{5.00}3dx+{\displaystyle {\int }_0^{0}4dy}}\right]\text{J}\\ \text{=}\left(\text{3}{\left[x\right]}_0^{5.00}+0\right)\text{J}\\ \text{=}\left(\text{3}\left[5.00-0\right]\right)\text{J}\\ \text{=15}\text{J}\end{array}$

Write the formula to calculate the work done by the force on the particle along the red path $OBC$

    $W_{\text{red}}=W_{OB}+W_{\text{BC}}$

Here, $W_{\text{red}}$ is the work done along the red path, $W_{\text{OB}}$ is the work done along the path $OB$ and $W_{\text{BC}}$ is the work done along the path $BC$.

Substitute $20\text{J}$ for $W_{\text{OB}}$ and $15\text{J}$ for $W_{\text{BC}}$.

    $\begin{array}{c}{W}_{\text{red}}=20+15\text{J}\\ =35\text{J}\end{array}$

Write the formula to calculate the work done by the force on the particle

    $W=\left[{\displaystyle \int 3dx+}{\displaystyle \int 4dy}\right]\text{J}$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00\text{m}$.

Taking the limits on integration,

    $\begin{array}{c}{W}_{\text{blue}}=\left[{\displaystyle {\int }_0^{5.00}3dx+}{\displaystyle {\int }_0^{5.00}4dx}\right]\text{J}\\ =\left(3{\left[x\right]}_0^{5.00}+4{\left[y\right]}_0^{5.00}\right)\text{J}\\ =\left(3\left[5.00-0\right]+4\left[5.00-0\right]\right)\text{J}\\ =\text{35}\text{J}\end{array}$

Since the work done by the force $\overrightarrow{\text{F}}=\left(3\widehat{\text{i}}\text{+4}\widehat{\text{j}}\right)\text{N}$ in all three purple, red and blue path is $35\text{J}$ hence the force is conservative and the work done is identical in all the three paths.

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the blue path is $\text{35}\text{J}$ and the work done by the force is same in all the three paths.

To determine

Whether the work done by the force $\overrightarrow{\text{F}}=\left(\text{4x}\widehat{\text{i}}\text{+3y}\widehat{\text{j}}\right)\text{N}$ on the particle along each one of the three paths shown in the figure is identical.

Answer to Problem 32P

The work done by the force $\overrightarrow{\text{F}}=\left(\text{4x}\widehat{\text{i}}\text{+3y}\widehat{\text{j}}\right)\text{N}$ on the particle along all the three path are $35\text{J}$ and the work done by the force is identical in all the three paths.

Explanation of Solution

The force acting on the particle is $\overrightarrow{\text{F}}=\left(\text{4x}\widehat{\text{i}}\text{+3y}\widehat{\text{j}}\right)\text{N}$ and the three different path of the particle moves is shown in figure (I).

Write the formula to calculate the work done by the force on the particle

    $W={\displaystyle \int \overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$

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

    $\begin{array}{c}W={\displaystyle \int \left(\left(\text{4x}\widehat{\text{i}}\text{+3y}\widehat{\text{j}}\right)\text{N}\right)\cdot \left(dx\widehat{\text{i}}\right)}\text{m}\\ ={\displaystyle \int \left(4xdx\right)\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}}\\ =\left[{\displaystyle \int 4xdx}\right]\text{J}\end{array}$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00\text{m}$ and there is no displacement in y-direction that is $y=0$.

Taking the limits of the integration,

    $\begin{array}{c}{W}_{\text{OA}}=\left[{\displaystyle {\int }_0^{5.00}4xdx}\right]\text{J}\\ =\left(4{\left[x\right]}_0^{5.00}\right)\text{J}\\ =\left(4\left[5.00-0\right]\right)\text{J}\\ =20\text{J}\end{array}$

Write the formula to calculate the work done by the force on the particle

    $W={\displaystyle \int \overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$

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

    $\begin{array}{c}W={\displaystyle \int \left(\left(\text{4x}\widehat{\text{i}}\text{+3y}\widehat{\text{j}}\right)\text{N}\right)\cdot \left(dy\widehat{\text{j}}\right)}\text{m}\\ ={\displaystyle \int \left(3ydy\right)\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}}\\ =\left[{\displaystyle \int 3ydy}\right]\text{J}\end{array}$

In the path $AC$, the particle moves from $y=0$ to $y=5.00\text{m}$ and there is a displacement of particle in x-direction that is $x=5.00\text{m}$.

Taking the limits of the integration,

    $\begin{array}{c}{W}_{\text{AC}}=\left[{\displaystyle {\int }_0^{5.00}3y\text{dy}}\right]\text{J}\\ \text{=}\left(3{\left[y\right]}_0^{5.00}\right)\text{J}\\ =3\left[5.00-0\right]\\ =15\text{J}\end{array}$

Write the formula to calculate the work done by the force on the particle along the purple path $OAC$

    $W_{\text{purple}}=W_{\text{OA}}+W_{\text{AC}}$                                 (III)

Here, $W_{\text{purple}}$ is the work done along the purple path, $W_{\text{OA}}$ is the work done along the path $OA$ and $W_{\text{AC}}$ is the work done along the path $AC$.

Substitute $20\text{J}$ for $W_{\text{OA}}$ and $15\text{J}$ for $W_{\text{AC}}$ in equation (II).

    $\begin{array}{c}{W}_{\text{purple}}=20\text{J+}15\text{J}\\ =35\text{J}\end{array}$

Write the formula to calculate the work done by the force on the particle

    $W={\displaystyle \int \overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$

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

    $\begin{array}{c}W={\displaystyle \int \left(\left(\text{4x}\widehat{\text{i}}\text{+3y}\widehat{\text{j}}\right)\text{N}\right)\cdot \left(dy\widehat{\text{j}}\right)}\text{m}\\ ={\displaystyle \int \left(3ydy\right)\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}}\\ =\left[{\displaystyle \int 3ydy}\right]\text{J}\end{array}$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00\text{m}$ and there is no displacement in x-direction that is $x=0$.

Taking the limits of the integration,

    $\begin{array}{c}{W}_{\text{OB}}=\left[{\displaystyle {\int }_0^{5.00}3ydy}\right]\text{J}\\ =3{\left[y\right]}_0^{5.00}\\ =\left(3\left[5.00-0\right]\right)\text{J}\\ =15\text{J}\end{array}$

Write the formula to calculate the work done by the force on the particle

    $W={\displaystyle \int \overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$

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

    $\begin{array}{c}W={\displaystyle \int \left(\left(\text{4x}\widehat{\text{i}}\text{+3y}\widehat{\text{j}}\right)\text{N}\right)\cdot \left(dx\widehat{\text{i}}\right)}\text{m}\\ ={\displaystyle \int \left(4xdx\right)\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}}\\ =\left[{\displaystyle \int 4xdx}\right]\text{J}\end{array}$

In the path $BC$, the particle moves from $x=0$ to $x=5.00\text{m}$ and there is a displacement of particle in y-direction that is $y=5.00\text{m}$.

Taking the limits of the integration,

    $\begin{array}{c}{W}_{\text{BC}}=\left[{\displaystyle {\int }_0^{5.00}4xdx}\right]\text{J}\\ \text{=}\left(4{\left[x\right]}_0^{5.00}\right)\text{J}\\ \text{=}\left(4\left[5.00-0\right]\right)\text{J}\\ =\text{20}\text{J}\end{array}$

Write the formula to calculate the work done by the force on the particle along the red path $OBC$

    $W_{\text{red}}=W_{OB}+W_{\text{BC}}$

Here, $W_{\text{red}}$ is the work done along the red path, $W_{\text{OB}}$ is the work done along the path $OB$ and $W_{\text{BC}}$ is the work done along the path $BC$.

Substitute $15\text{J}$ for $W_{\text{OB}}$ and $20\text{J}$ for $W_{\text{BC}}$.

    $\begin{array}{c}{W}_{\text{red}}=15\text{J+}20\text{J}\\ =35\text{J}\end{array}$

Write the formula to calculate the work done by the force on the particle

    $W=\left[{\displaystyle \int 4xdx+}{\displaystyle \int 3ydy}\right]\text{J}$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00\text{m}$.

Taking the limits on integration,

    $\begin{array}{c}{W}_{\text{red}}=\left[{\displaystyle {\int }_0^{5.00}4xdx+}{\displaystyle {\int }_0^{5.00}3ydx}\right]\text{J}\\ =\left(4{\left[x\right]}_0^{5.00}+3{\left[y\right]}_0^{5.00}\right)\text{J}\\ =\left(4\left[5.00-0\right]+3\left[5.00-0\right]\right)\text{J}\\ =\text{35}\text{J}\end{array}$

Since the work done by the force $\overrightarrow{\text{F}}=\left(3\widehat{\text{i}}\text{+4}\widehat{\text{j}}\right)\text{N}$ in all three purple, red and blue path is $35\text{J}$ hence the force is conservative and the work done is identical in all the three paths.

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the blue path is $\text{35}\text{J}$ and the work done by the force is same in all the three paths.

To determine

Whether the work done by the force $\overrightarrow{\text{F}}=\left(\text{y}\widehat{\text{i}}-\text{x}\widehat{\text{j}}\right)\text{N}$ on the particle along each one of the three paths shown in the figure is identical.

Answer to Problem 32P

The work done by the force $\overrightarrow{\text{F}}=\left(y\widehat{\text{i}}-x\widehat{\text{j}}\right)\text{N}$ on the particle along all the three path are not identical.

Explanation of Solution

Write the formula to calculate the work done by the force on the particle

    $W={\displaystyle \int \overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$

Substitute $\left(y\widehat{\text{i}}-x\widehat{\text{j}}\right)\text{N}$ for $\overrightarrow{\text{F}}$ and $dx\widehat{\text{i}}$ for $d\widehat{\text{r}}$.

    $\begin{array}{c}W={\displaystyle \int \left(\left(y\widehat{\text{i}}-x\widehat{\text{j}}\right)\text{N}\right)\cdot \left(dx\widehat{\text{i}}\right)}\text{m}\\ ={\displaystyle \int \left(ydx\right)\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}}\\ =\left[{\displaystyle \int ydx}\right]\text{J}\end{array}$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00\text{m}$ and there is no displacement in y-direction that is $y=0$.

Substitute $0$ for $y$ and $0$ for $dy$.

    $\begin{array}{c}{W}_{\text{OA}}=\left[{\displaystyle \int 0dx}\right]\text{J}\\ =\text{0}\text{J}\end{array}$

Write the formula to calculate the work done by the force on the particle

    $W={\displaystyle \int \overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$

Substitute $\left(y\widehat{\text{i}}-x\widehat{\text{j}}\right)\text{N}$ for $\overrightarrow{\text{F}}$ and $dy\widehat{\text{j}}$ for $d\widehat{\text{r}}$.

    $\begin{array}{c}W={\displaystyle \int \left(\left(y\widehat{\text{i}}-x\widehat{\text{j}}\right)\text{N}\right)\cdot \left(dy\text{j}\right)}\text{m}\\ =-{\displaystyle \int \left(xdy\right)\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}}\\ =-\left[{\displaystyle \int xdy}\right]\text{J}\end{array}$

In the path $AC$, the particle moves from $y=0$ to $y=5.00\text{m}$ and there is a displacement of particle in x-direction that is $x=5.00\text{m}$.

Substitute $5.00$ for $x$ and $0$ for $dx$.

    $W_{AC}=\left[-{\displaystyle \int \left(5.00\right)dy}\right]\text{J}$

Taking the limits of the integration,

    $\begin{array}{c}{W}_{\text{AC}}=\left[{\displaystyle {\int }_0^{5.00}\left(5.00\right)dy}\right]\text{J}\\ \text{=}\left(-5{\left[y\right]}_0^{5.00}\right)\text{J}\\ =-5\left[5.00-0\right]\\ =-25\text{J}\end{array}$

Write the formula to calculate the work done by the force on the particle along the purple path $OAC$

    $W_{\text{purple}}=W_{\text{OA}}+W_{\text{AC}}$

Here, $W_{\text{purple}}$ is the work done along the purple path, $W_{\text{OA}}$ is the work done along the path $OA$ and $W_{\text{AC}}$ is the work done along the path $AC$.

Substitute $0$ for $W_{\text{OA}}$ and $-25\text{J}$ for $W_{\text{AC}}$.

    $\begin{array}{c}{W}_{\text{purple}}=0\text{J}-25\text{J}\\ =-25\text{J}\end{array}$

Write the formula to calculate the work done by the force on the particle

    $W={\displaystyle \int \overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$

Substitute $\left(y\widehat{\text{i}}-x\widehat{\text{j}}\right)\text{N}$ for $\overrightarrow{\text{F}}$ and $dy\widehat{\text{j}}$ for $d\widehat{\text{r}}$.

    $\begin{array}{c}W={\displaystyle \int \left(\left(y\widehat{\text{i}}-x\widehat{\text{j}}\right)\text{N}\right)\cdot \left(dy\text{j}\right)}\text{m}\\ =-{\displaystyle \int \left(xdy\right)\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}}\\ =-\left[{\displaystyle \int xdy}\right]\text{J}\end{array}$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00\text{m}$ and there is no displacement in x-direction that is $x=0$.

Substitute $0$ for $x$ in above integration.

    $\begin{array}{c}{W}_{\text{OB}}=\left[{\displaystyle {\int }_0^{5.00}0dy}\right]\text{J}\\ =0\end{array}$

Write the formula to calculate the work done by the force on the particle

    $W={\displaystyle \int \overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$

Substitute $\left(y\widehat{\text{i}}-x\widehat{\text{j}}\right)\text{N}$ for $\overrightarrow{\text{F}}$ and $dx\widehat{\text{i}}$ for $d\widehat{\text{r}}$.

    $\begin{array}{c}W={\displaystyle \int \left(\left(y\widehat{\text{i}}-x\widehat{\text{j}}\right)\text{N}\right)\cdot \left(dx\widehat{\text{i}}\right)}\text{m}\\ ={\displaystyle \int \left(ydx\right)\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}}\\ =\left[{\displaystyle \int ydx}\right]\text{J}\end{array}$

In the path $BC$, the particle moves from $x=0$ to $x=5.00\text{m}$ and there is a displacement of particle in y-direction that is $y=5.00\text{m}$.

Substitute $5.00$ for $y$ and $0$ for $dy$.

Taking the limits of the integration,

    $\begin{array}{c}{W}_{\text{BC}}=\left[{\displaystyle {\int }_0^{5.00}5dx}\right]\text{J}\\ \text{=}\left(5{\left[x\right]}_0^{5.00}\right)\text{J}\\ \text{=}\left(5\left[5.00-0\right]\right)\text{J}\\ =\text{25}\text{J}\end{array}$

Write the formula to calculate the work done by the force on the particle along the red path $OBC$

    $W_{\text{red}}=W_{OB}+W_{\text{BC}}$                                 (VI)

Here, $W_{\text{red}}$ is the work done along the red path, $W_{\text{OB}}$ is the work done along the path $OB$ and $W_{\text{BC}}$ is the work done along the path $BC$.

Substitute $0$ for $W_{\text{OB}}$ and $25\text{J}$ for $W_{\text{BC}}$ in equation (VI).

    $\begin{array}{c}{W}_{\text{red}}=0\text{J}+25\text{J}\\ =25\text{J}\end{array}$

Write the formula to calculate the work done by the force on the particle

    $W={\displaystyle \int \overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$

Substitute $\left(y\widehat{\text{i}}-x\widehat{\text{j}}\right)\text{N}$ for $\overrightarrow{\text{F}}$ and $dx\widehat{\text{i}}\text{+}dy\widehat{\text{j}}$ for $d\widehat{\text{r}}$.

    $\begin{array}{c}W={\displaystyle \int \left(\left(y\widehat{\text{i}}-x\widehat{\text{j}}\right)\text{N}\right)\cdot \left(dx\widehat{\text{i}}\text{+}dy\widehat{\text{j}}\right)}\text{m}\\ ={\displaystyle \int \left(ydx-xdy\right)\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}}\\ =\left[{\displaystyle \int ydx-xdy}\right]\text{J}\end{array}$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00\text{m}$. Hence equation of line become $y=x$.

Substitute $x$ for $y$ and $dx$ for $dy$.

Taking the limits on integration,

    $\begin{array}{c}{W}_{\text{blue}}=\left[{\displaystyle \int xdx-{\displaystyle \int xdx}}\right]\text{J}\\ =0\end{array}$

Conclusion:

Therefore, the work done by the force on the particle as it goes along the three paths is not same.