Chapter 9, Problem 30PQ

A particle moves in the xy plane (Fig. P9.30) from the origin to a point having coordinates x = 7.00 m and y = 4.00 m under the influence of a force given by $\stackrel{\to }{F}=3y^2î+x�$.

1. a. What is the work done on the particle by the force F if it moves along path 1 (shown in red)?
2. b. What is the work done on the particle by the force F if it moves along path 2 (shown in blue)?
3. c. What is the work done on the particle by the force F if it moves along path 3 (shown in green)?
4. d. Is the force F conservative or nonconservative? Explain.

FIGURE P9.30

In each case, the work is found using the integral of $\stackrel{\to }{F}·d\stackrel{\to }{r}$ along the path (Equation 9.21).

$W={\displaystyle {\int }_{r_t}^{r_f}\stackrel{\to }{F}·d\stackrel{\to }{r}}={\displaystyle {\int }_{r_t}^{r_f}(F_{x}dx+F_{y}dy+F_{z}dz)}$

(a) The work done along path 1, we first need to integrate along $d\stackrel{\to }{r}=dx\hat{i}$ from (0,0) to (7,0) and then along $d\stackrel{\to }{r}=dy\hat{j}$ from (7,0) to (7,4):

$W_1={\displaystyle \underset{x=0;y=0}{\overset{x=7;y=0}{\int }}(3y^2\hat{i}+x\hat{j})·(dx\hat{i})}+{\displaystyle \underset{x=7;y=0}{\overset{x=7;y=4}{\int }}(3y^2\hat{i}+x\hat{j})·(dy\hat{j})}$

Performing the dot products, we get

$W_1={\displaystyle \underset{x=0;y=0}{\overset{x=7;y=0}{\int }}3y^{2}dx}+{\displaystyle \underset{x=7;y=0}{\overset{x=7;y=4}{\int }}xdy}$

Along the first part of this path, y = 0 therefore the first integral equals zero. For the second integral, x is constant and can be pulled out of the integral, and we can evaluate dy.

$W_1=0+{\displaystyle \underset{x=7;y=0}{\overset{x=7;y=4}{\int }}{xdy=xy|}_{x=7;y=0}^{x=7;y=4}=\boxed{28\text{ J}}}$

(b) The work done along path 2 is along $d\stackrel{\to }{r}=dy\hat{j}$ from (0,0) to (0,4) and then along $d\stackrel{\to }{r}=dx\hat{i}$ from (0,4) to (7,4):

$W_2={\displaystyle \underset{x=0;y=0}{\overset{x=0;y=4}{\int }}(3y^2\hat{i}+x\hat{j})·(dy\hat{j})}+{\displaystyle \underset{x=0;y=4}{\overset{x=7;y=4}{\int }}(3y^2\hat{i}+x\hat{j})·(dy\hat{i})}$

Performing the dot product, we get:

$W_2={\displaystyle \underset{x=0;y=0}{\overset{x=0;y=4}{\int }}xdy+}{\displaystyle \underset{x=0;y=4}{\overset{x=7;y=4}{\int }}3y^{2}dx}$

Along the first part of this path, x = 0. Therefore, the first integral equals zero. For the second integral, y is constant and can be pulled out of the integral, and we can evaluate dx.

$W_2={0+3y^{2}x|}_{x=0;y=4}^{x=7;y=4}=\boxed{336\text{ J}}$

(c) To find the work along the third path, we first write the expression for the work integral.

$\begin{array}{l}W={\displaystyle {\int }_{r_t}^{r_f}\stackrel{\to }{F}·d\stackrel{\to }{r}}={\displaystyle {\int }_{r_t}^{r_f}(F_{x}dx+F_{y}dy+F_{z}dz)}\\ W={\displaystyle {\int }_{r_t}^{r_f}(3y^{2}dx+xdy)}\end{array}$    (1)

At first glance, this appears quite simple, but we can’t integrate ${\displaystyle \int xdy=xy}$ like we might have above because the value of x changes as we vary y (i.e., x is a function of y.) [In parts (a) and (b), on a straight horizontal or vertical line, only x or y changes]. One approach is to parameterize both x and y as a function of another variable, say t, and write each integral in terms of only x or y. Constraining dr to be along the desired line, we can relate dx and dy:

$\begin{array}{l}\tan \theta =\frac{dy}{dx}\\ dy=\tan \theta dx\text{ and }dx=\frac{dy}{\tan \theta }\end{array}$    (2)

Now, use equation (2) in (1) to express each integral in terms of only one variable.

$\begin{array}{l}W={\displaystyle \underset{x=0;y=0}{\overset{x=7;y=4}{\int }}3y^{2}dx}+{\displaystyle \underset{x=0;y=0}{\overset{x=7;y=4}{\int }}xdy}\\ W={\displaystyle \underset{y=0}{\overset{y=4}{\int }}3y^2\frac{dy}{\tan \theta }}+{\displaystyle \underset{x=0}{\overset{x=7}{\int }}x\tan \theta dx}\end{array}$

We can determine the tangent of the angle, which is constant (the angle is the angle of the line with respect to the horizontal).

$\tan \theta =\frac{4.00}{7.00}=0.570$

Insert the value of the tangent and solve the integrals.

$\begin{array}{l}W=\frac{3}{0.570}{\frac{y^3}{3}|}_{y=0}^{y=4}+{0.570\frac{x^2}{2}|}_{x=0}^{x=7}\\ W=112+14=\boxed{126\text{ J}}\end{array}$

(d) Since the work done is not “path-independent”, this is $\boxed{\text{non-conservative}}$ force.

Figure P9.30ANS

To determine

The work done on the particle by the force $F$ if it moves along the path 1.

Answer to Problem 30PQ

The work done on the particle by the force $F$ if it moves along the path 1 is $\underset{\_}{28\text{J}}$.

Explanation of Solution

The path 1 followed by the particle is given in Figure P9.30. Given that the coordinate of the final position of the particle is $\left(7.00,4.00\right)$ and the force is $\overrightarrow{\text{F}}=3y^2\widehat{\text{i}}+x\widehat{\text{j}}$.

Write the expression for the work done by a force.

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

Here, $W$ is the work done, $r_f$ is the final position, $r_i$ is the initial position, and $d\overrightarrow{\text{r}}$ is the elemental path.

The path 1 of the particle consist of two parts. Motion from $\left(0,0\right)$ to $\left(7,0\right)$ and $\left(7,0\right)$ to $\left(7,4\right)$. Thus, the integration has to be performed along $d\overrightarrow{\text{r}}=dx\widehat{\text{i}}$ from $\left(0,0\right)$ to $\left(7,0\right)$ and then along $d\overrightarrow{\text{r}}=dy\widehat{\text{j}}$ from $\left(7,0\right)$ to $\left(7,4\right)$.

Use the force vector along with the limits of integration and perform the integration (represent the work done along path 1 as $W_1$).

  $W_1={\displaystyle \underset{x=0;y=0}{\overset{x=7;y=0}{\int }}\left(3y^2\widehat{\text{i}}+x\widehat{\text{j}}\right)·\left(dx\widehat{\text{i}}\right)}+{\displaystyle \underset{x=7;y=0}{\overset{x=7;y=4}{\int }}\left(3y^2\widehat{\text{i}}+x\widehat{\text{j}}\right)·\left(dy\widehat{\text{j}}\right)}$                                        (II)

Perform the dot product to reduce the integral (II).

  $W_1={\displaystyle \underset{x=0;y=0}{\overset{x=7;y=0}{\int }}3y^{2}dx}+{\displaystyle \underset{x=7;y=0}{\overset{x=7;y=4}{\int }}xdy}$                                                                        (III)

Along the first part of the path 1, $y=0$, and hence the first integral equals zero. For the second integral, $x$ is constant and can be pulled out of the integral, and $dy$ can be evaluated.

  $\begin{array}{c}{W}_1=0+{\displaystyle \underset{x=7;y=0}{\overset{x=7;y=4}{\int }}xdy}\\ ={xy|}_{x=7;y=0}^{x=7;y=4}\\ =28-0\\ =28\text{J}\end{array}$

Conclusion:

Therefore, the work done on the particle by the force $F$ if it moves along the path 1 is $\underset{\_}{28\text{J}}$.

To determine

The work done on the particle by the force $F$ if it moves along the path 2.

Answer to Problem 30PQ

The work done on the particle by the force $F$ if it moves along the path 2 is $\underset{\_}{336\text{J}}$.

Explanation of Solution

The path 2 followed by the particle is given in Figure P9.30. Given that the coordinate of the final position of the particle is $\left(7.00,4.00\right)$ and the force is $\overrightarrow{\text{F}}=3y^2\widehat{\text{i}}+x\widehat{\text{j}}$.

Equation (I) gives the expression for the work done by a force.

  $W={\displaystyle \underset{r_i}{\overset{r_f}{\int }}\overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$

The path 2 of the particle consist of two parts. Motion from $\left(0,0\right)$ to $\left(0,4\right)$ and $\left(0,4\right)$ to $\left(7,4\right)$. Thus, the integration has to be performed along $d\overrightarrow{\text{r}}=dy\widehat{\text{j}}$ from $\left(0,0\right)$ to $\left(0,4\right)$ and then along $d\overrightarrow{\text{r}}=dx\widehat{\text{i}}$ from $\left(0,4\right)$ to $\left(7,4\right)$.

Use the force vector along with the limits of integration and perform the integration (represent the work done along path 2 as $W_2$).

  $W_2={\displaystyle \underset{x=0;y=0}{\overset{x=0;y=4}{\int }}\left(3y^2\widehat{\text{i}}+x\widehat{\text{j}}\right)·\left(dy\widehat{\text{j}}\right)}+{\displaystyle \underset{x=0;y=4}{\overset{x=7;y=4}{\int }}\left(3y^2\widehat{\text{i}}+x\widehat{\text{j}}\right)·\left(dx\widehat{\text{i}}\right)}$                                     (IV)

Perform the dot product to reduce the integral (IV).

  $W_2={\displaystyle \underset{x=0;y=0}{\overset{x=0;y=4}{\int }}xdy}+{\displaystyle \underset{x=0;y=4}{\overset{x=7;y=4}{\int }}3y^{2}dx}$                                                                         (V)

Along the first part of the path 2, $x=0$, and hence the first integral equals zero. For the second integral, $y$ is constant and can be pulled out of the integral, and $dx$ can be evaluated.

  $\begin{array}{c}{W}_2=0+{\displaystyle \underset{x=0;y=4}{\overset{x=7;y=4}{\int }}3y^{2}dx}\\ ={3y^{2}x|}_{x=0;y=4}^{x=7;y=4}\\ =336-0\\ =336\text{J}\end{array}$

Conclusion:

Therefore, the work done on the particle by the force $F$ if it moves along the path 2 is $\underset{\_}{336\text{J}}$.

To determine

The work done on the particle by the force $F$ if it moves along the path 3.

Answer to Problem 30PQ

The work done on the particle by the force $F$ if it moves along the path 3 is $\underset{\_}{126\text{J}}$.

Explanation of Solution

The path 3 followed by the particle is given in Figure P9.30. Given that the coordinate of the final position of the particle is $\left(7.00,4.00\right)$ and the force is $\overrightarrow{\text{F}}=3y^2\widehat{\text{i}}+x\widehat{\text{j}}$.

Equation (I) gives the expression for the work done by a force.

  $W={\displaystyle \underset{r_i}{\overset{r_f}{\int }}\overrightarrow{\text{F}}\cdot d\overrightarrow{\text{r}}}$

Write equation (I) in terms of $x$, $y$, and $z$ components.

  $W={\displaystyle \underset{r_i}{\overset{r_f}{\int }}\left(F_{x}dx+F_{y}dy+F_{z}dz\right)}$                                                                               (VI)

Use the $x$ and $y$ components of the given force in the integral (VI). Since the motion is confined in $xy$ plane, there is no $z$ component in the integral.

  $W={\displaystyle \underset{r_i}{\overset{r_f}{\int }}\left(3y^{2}dx+xdy\right)}$                                                                                        (VII)

The path 3 of the particle starts from $\left(0,0\right)$ and ends at $\left(7,4\right)$. Thus, the integral (VII) can be written as (represent the work done along path 3 as $W_3$),

  $W_3={\displaystyle \underset{x=0;y=0}{\overset{x=7;y=4}{\int }}3y^{2}dx}+{\displaystyle \underset{x=0;y=0}{\overset{x=7;y=4}{\int }}xdy}$                                                                         (VIII)

Here, both $x$ and $y$ coordinates change and hence direct integration cannot be performed. Let us consider the angle $\theta$ that the vector of path $3$ make with the $x$ axis as shown in Figure 1.

Write the expression relating $dx$, $dy$, and $\theta$.

  $\tan \theta =\frac{dy}{dx}$                                                                                                    (IX)

Solve equation (IX) for $dy$.

  $dy=\tan \theta dx$                                                                                                     (X)

Solve equation (IX) for $dx$.

  $dx=\frac{dy}{\tan \theta }$                                                                                                       (XI)

Use equation (X) and (XI) in (VIII).

  $W_3={\displaystyle \underset{x=0;y=0}{\overset{x=7;y=4}{\int }}3y^2\frac{dy}{\tan \theta }}+{\displaystyle \underset{x=0;y=0}{\overset{x=7;y=4}{\int }}x\tan \theta dx}$                                                          (XII)

Compute $\tan \theta$ from the paths shown in Figure P9.30.

  $\begin{array}{c}\tan \theta =\frac{4.00}{7.00}\\ =0.570\end{array}$                                                                                             (XIII)

Use equation (XIII) in (XII) and perform the integral.

  $\begin{array}{c}{W}_3={\displaystyle \underset{y=0}{\overset{y=4}{\int }}\frac{3}{0.570}{y}^{2}dy}+{\displaystyle \underset{x=0}{\overset{x=7}{\int }}0.570xdx}\\ =\frac{3}{0.570}{\frac{y^3}{3}|}_{y=0}^{y=4}+0.570{\frac{x^2}{2}|}_{x=0}^{x=7}\\ =112+14\\ =126\text{J}\end{array}$

Conclusion:

Therefore, the work done on the particle by the force $F$ if it moves along the path 3 is $\underset{\_}{126\text{J}}$.

To determine

Whether the force $\overrightarrow{\text{F}}$ is conservative or non-conservative.

Answer to Problem 30PQ

The force $\overrightarrow{\text{F}}$ is non-conservative in nature.

Explanation of Solution

From part (a), (b) and (c) it is found that the work done by the force $\overrightarrow{\text{F}}$ on moving the particle from $\left(0,0\right)$ to $\left(7,4\right)$ along different path is different. By the definition of conservative forces, the work done by a conservative force is path independent. However, here the work done by the force $\overrightarrow{\text{F}}$ is path dependent and hence it is a non-conservative force.

Conclusion:

Therefore, the force $\overrightarrow{\text{F}}$ is non-conservative in nature.