Chapter 6, Problem 44P

(a)

To determine

To Show: That the constant force act on the object is conservative.

Answer to Problem 44P

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.

Given Information:

The general definition for work done by a force is,

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

Formula to calculate the work done by the force on the object is,

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

• $W$ is the work done by the force on the particle.
• $\overrightarrow{\text{F}}$ is the applied force vector on the object.
• $d\overrightarrow{\text{r}}$ 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.

(b)

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 44P

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

Section-1:

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 the path $OA$.

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $35\text{J}$.

Given Information:

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)

Formula to calculate the work done by the force on the particle is,

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

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,

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

Section-2:

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 the path $AC$.

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $35\text{J}$.

Given Information:

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).

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 $x=5.00\text{m}$.

Formula to calculate the work done by the force on the particle is,

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

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,

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

Section-3:

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 the purple path.

Answer: The work done by the force on the particle as it goes from O to C along the purple path is $35\text{J}$.

Given Information:

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).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

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

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

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

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

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $35\text{J}$.

Section-4:

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 the path $OB$.

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $35\text{J}$.

Given Information:

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).

Formula to calculate the work done by the force on the particle is,

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

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,

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

Section-5:

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 the path $BC$.

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $35\text{J}$.

Given Information:

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).

Formula to calculate the work done by the force on the particle is,

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

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,

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

Section-6:

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 the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $35\text{J}$.

Given Information:

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).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

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

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

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

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

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $35\text{J}$.

Section-7:

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 the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $35\text{J}$.

Given Information:

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).

Formula to calculate the work done by the force on the particle is,

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

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.

(c)

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 44P

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

Section-1:

To determine: 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 the path $OA$.

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $35\text{J}$.

Given Information:

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).

Formula to calculate the work done by the force on the particle is,

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

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,

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

Section-2:

To determine: 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 the path $AC$.

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $35\text{J}$.

Given Information:

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).

Formula to calculate the work done by the force on the particle is,

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

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,

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

Section-3:

To determine: 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 the purple path.

Answer: The work done by the force on the particle as it goes from O to C along the purple path is $35\text{J}$.

Given Information:

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).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

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

• $W_{\text{purple}}$ is the work done along the purple path.
• $W_{\text{OA}}$ is the work done along the path $OA$.
• $W_{\text{AC}}$ 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).

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

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $35\text{J}$.

Section-4:

To determine: 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 the path $OB$.

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $35\text{J}$.

Given Information:

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).

Formula to calculate the work done by the force on the particle is,

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

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,

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

Section-5:

To determine: 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 the path $BC$.

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $20\text{J}$.

Given Information:

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).

Formula to calculate the work done by the force on the particle is,

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

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,

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

Section-6:

To determine: 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 the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $35\text{J}$.

Given Information:

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).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

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

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

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

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

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $35\text{J}$.

Section-7:

To determine: 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 the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $35\text{J}$.

Given Information:

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).

Formula to calculate the work done by the force on the particle is,

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

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.

(d)

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 44P

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

Section-1:

To determine: 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 the path $OA$.

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $-25\text{J}$.

Given Information:

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

Formula to calculate the work done by the force on the particle is,

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

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 $dy$.

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

Section-2:

To determine: 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 the path $AC$.

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $-25\text{J}$.

Given Information:

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

Formula to calculate the work done by the force on the particle is,

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

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,

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

Section-3:

To determine: 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 the path $AC$.

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $-25\text{J}$.

Given Information:

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

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

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

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

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

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

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $-25\text{J}$.

Section-4:

To determine: 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 the path $OB$.

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $25\text{J}$.

Given Information:

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

Formula to calculate the work done by the force on the particle is,

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

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.

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

Section-5:

To determine: 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 the path $BC$.

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $25\text{J}$.

Given Information:

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

Formula to calculate the work done by the force on the particle is,

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

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,

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

Section-6:

To determine: 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 the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path  is $25\text{J}$.

Given Information:

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

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

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

• $W_{\text{red}}$ is the work done along the red path.
• $W_{\text{OB}}$ is the work done along the path $OB$.
• $W_{\text{BC}}$ 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).

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

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $25\text{J}$.

Section-7:

To determine: 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 the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $0$.

Given Information:

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

Formula to calculate the work done by the force on the particle is,

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

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.