Chapter 7, Problem 33P

A force acting on a particle moving in the xy plane is given by $\stackrel{\to }{F}=(2y\hat{i}+x^2\hat{j})$, where $\stackrel{\to }{F}$ is in newtons and x and y are in meters. The particle moves from the origin to a final position having coordinates x = 5.00 m and y = 5.00 m as shown in Figure P7.31. Calculate the work done by $\stackrel{\to }{F}$ on the particle as it moves along (a) the purple path, (b) the red path, and (c) the blue path, (d) Is $\stackrel{\to }{F}$ conservative or nonconservative? (e) Explain your answer to part (d).

Figure P7.31

To determine

The amount of work done by the force along the purple path.

Answer to Problem 33P

The work done by the force along the purple path is $\text{125}\text{J}$.

Explanation of Solution

Given info: The coordinates of point (C) is $\left(5.00\text{m, 5}\text{.00}\text{m}\right)$ and the force acting on the body is $\overrightarrow{F}=2y\widehat{\text{i}}+x^2\widehat{\text{j}}$.

Along path OA the work is done is only in $x$ direction so, work done along OA is,

$W_{\text{OA}}={\displaystyle {\int }_0^{5.00}\overrightarrow{F}}\cdot dx\widehat{\text{i}}$

Here,

$F$ is the force acting on the particle.

Substitute $2y\widehat{\text{i}}+x^2\widehat{\text{j}}$ for $F$ in the above equation.

$\begin{array}{c}{W}_{\text{OA}}={\displaystyle {\int }_0^{5.00}\left(2y\widehat{\text{i}}+x^2\widehat{\text{j}}\right)}\cdot dx\widehat{\text{i}}\\ \text{=}{\displaystyle {\int }_0^{5.00}\left(2y\right)}\cdot dx\end{array}$

The value $y$ along path OA is $0$.

Substitute $0$ for $y$ in the above equation.

$\begin{array}{c}{W}_{\text{OA}}\text{=}{\displaystyle {\int }_0^{5.00}\left(2\left(0\right)\right)}\cdot dx\\ =0\text{J}\end{array}$

Thus, work done along OA is $0\text{J}$.

Work done along AC is,

$W_{\text{AC}}={\displaystyle {\int }_0^{5.00}\overrightarrow{F}}\cdot dy\widehat{\text{j}}$

Substitute $2y\widehat{\text{i}}+x^2\widehat{\text{j}}$ for $F$ in the above equation.

$\begin{array}{c}{W}_{\text{AC}}={\displaystyle {\int }_0^{5.00}\left(2y\widehat{\text{i}}+x^2\widehat{\text{j}}\right)}\cdot dy\widehat{\text{j}}\\ \text{=}{\displaystyle {\int }_0^{5.00}\left(x^2\right)}\cdot dy\\ =x^2{\left[y\right]}_0^{5.00}\\ =5x^2\text{J}\end{array}$

The value of $x$ along the path AC is $5.00$.

Substitute $5.00$ for $x$ in the above equation.

$\begin{array}{c}{W}_{\text{AC}}=5\left({5.00}^2\right)\text{J}\\ \text{=125}\text{J}\end{array}$

Thus, the work done along AC is $125\text{J}$.

The work done along purple path is,

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

Substitute $0\text{J}$ for $W_{\text{OA}}$ and $125\text{J}$ for $W_{\text{AC}}$ in the above equation.

$\begin{array}{c}{W}_{\text{purple}}=0\text{J}+125\text{J}\\ =\text{125}\text{J}\end{array}$

Conclusion:

Therefore, the work done in the purple path is $125\text{J}$.

To determine

The amount of work done by the force along the red path.

Answer to Problem 33P

The work done by the force along the red path is $50.0\text{J}$.

Explanation of Solution

Given info: The coordinates of point (C) is $\left(5.00\text{m, 5}\text{.00}\text{m}\right)$ and the force acting on the body is $\overrightarrow{F}=2y\widehat{\text{i}}+x^2\widehat{\text{j}}$.

Work done along OB is,

$W_{\text{OB}}={\displaystyle {\int }_0^{5.00}\overrightarrow{F}}\cdot dy\widehat{\text{i}}$

Substitute $2y\widehat{\text{i}}+x^2\widehat{\text{j}}$ for $F$ in the above equation.

$\begin{array}{c}{W}_{\text{OB}}={\displaystyle {\int }_0^{5.00}\left(2y\widehat{\text{i}}+x^2\widehat{\text{j}}\right)}\cdot dy\widehat{\text{j}}\\ \text{=}{\displaystyle {\int }_0^{5.00}\left(x^2\right)}\cdot dy\end{array}$

The value of $x$ along path (OB) is zero.

Substitute $0$ for $x$ in the above equation.

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

Thus, the work done along the path OB is $0\text{J}$.

Work done along BC is,

$W_{\text{BC}}={\displaystyle {\int }_0^{5.00}\overrightarrow{F}}\cdot dx\widehat{\text{i}}$

Substitute $2y\widehat{\text{i}}+x^2\widehat{\text{j}}$ for $F$ in the above equation.

$\begin{array}{c}{W}_{\text{BC}}={\displaystyle {\int }_0^{5.00}\left(2y\widehat{\text{i}}+x^2\widehat{\text{j}}\right)}\cdot dx\widehat{\text{i}}\\ \text{=}{\displaystyle {\int }_0^{5.00}\left(2y\right)}\cdot dx\\ =2y{\left[x\right]}_0^{5.00}\\ =10y\text{J}\end{array}$

Along path BC the value of $y$ is $5.00$.

Substitute $5.00$ for $y$ in the above equation.

$\begin{array}{c}{W}_{\text{BC}}=10\left(5.00\right)\text{J}\\ =\text{50}\text{.0}\text{J}\end{array}$

Thus, the work done along the path OB is $50.0\text{J}$.

The work done along red path is,

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

Substitute $50.0\text{J}$ for $W_{\text{BC}}$ and $0\text{J}$ for $W_{\text{OB}}$ in the above equation.

$\begin{array}{c}{W}_{\text{red}}=0\text{J}+50.0\text{J}\\ =\text{50}\text{.0}\text{J}\end{array}$

Conclusion:

Therefore, the work done by the force along the red path is $50.0\text{J}$.

To determine

The amount of work done by the force along the blue path.

Answer to Problem 33P

The work done by the force along the blue path is $66.7\text{J}$.

Explanation of Solution

Given info: The coordinates of point (C) is $\left(5.00\text{m, 5}\text{.00}\text{m}\right)$ and the force acting on the body is $\overrightarrow{F}=2y\widehat{\text{i}}+x^2\widehat{\text{j}}$.

The work done along the blue path is,

$W_{\text{OC}}={\displaystyle {\int }_0^{5.00}\overrightarrow{F}}\cdot \left(dx\widehat{\text{i}}+dy\widehat{\text{i}}\right)$

Substitute $2y\widehat{\text{i}}+x^2\widehat{\text{j}}$ for $\overrightarrow{F}$ in the above equation

$W_{\text{OC}}={\displaystyle {\int }_0^{5.00}2y\widehat{\text{i}}+x^2\widehat{\text{j}}}\cdot \left(dx\widehat{\text{i}}+dy\widehat{\text{i}}\right)$

The value of $x$ and $y$ along path OC is same, so substitute $x$ for $y$ in the above equation.

$\begin{array}{c}{W}_{\text{OC}}={\displaystyle {\int }_0^{5.00}2x\widehat{\text{i}}+x^2\widehat{\text{j}}}\cdot \left(dx\right)\\ ={\left[x^2+\frac{x^3}{3}\right]}_0^{5.00}\\ =66.66\text{J}\\ \approx 66.7\text{J}\end{array}$

Conclusion:

Therefore, the work done by the force along the blue path is $66.7\text{J}$.

To determine

Whether $\overrightarrow{F}$ is conservative.

Answer to Problem 33P

The force $\overrightarrow{F}$ is not a conservative force.

Explanation of Solution

Given info: The coordinates of point (C) is $\left(5.00\text{m, 5}\text{.00}\text{m}\right)$ and the force acting on the body is $\overrightarrow{F}=2y\widehat{\text{i}}+x^2\widehat{\text{j}}$.

The work done by the force is non conservative in nature as a force is a conservative force if the work done is independent of the path the particle takes.

For the given case the work done in every case is different as the path changes for each of the cases. From the above calculation the work done by the force $\overrightarrow{F}$ is different for each path taken. So, the force $\overrightarrow{F}$ is non-conservative in nature.

Conclusion:

Therefore, the work done by force along every path is different hence it is a non conservative force in nature.

To determine

The reason for the force $\overrightarrow{F}$ is not a conservative force.

Answer to Problem 33P

The work done along each path is different.

Explanation of Solution

Given info: The coordinates of point (C) is $\left(5.00\text{m, 5}\text{.00}\text{m}\right)$ and the force acting on the body is $\overrightarrow{F}=2y\widehat{\text{i}}+x^2\widehat{\text{j}}$.

The work done by the force $\overrightarrow{F}$ along the purple path is $125\text{J}$, the work done along the red path is $50.0\text{J}$ and the work done along the blue path is $66.7\text{J}$.

The work done by the given force $\overrightarrow{F}$ is different for each path the particle traversed. The work done by a conservative force is independent of the path taken. So the force is non conservative in nature.

Conclusion:

Therefore, the force is non conservative in nature because work done along each path is different.