Question

Find the work done by the force $\mathit{F}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathit{y}\mathit{i}\mathbf{-}\mathit{x}{\mathit{y}}^{\mathbf{2}}\mathit{j}$ along the paths shown from (1,1) to (4,2).

Short answer

The total work done by force $F=x^{2}yi-x{y}^{2}j$ is $\frac{86}{3}$ and along the path from (1,1) to (4,2) is $\frac{-31}{3}$.

Step-by-step solution

Given Information

The given force vector is $F=x^{2}yi-x{y}^{2}j$ and path from (1,1) to (4,2) .

Definition of work done

Work done by a force is defined as the product of displacement of an object and component of force applied which is in the direction of displacement of an object.

Concept and Formula

The formula to find the work done by force is given by equation mentioned below.

$W=\int F.dr$

where F: force applied anddisplacement of an object.

Also, the equation of line joining two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given below.

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$

Calculate the total work done

Use the formula $W=\int F.dr$

Put $F=x^{2}yi-x{y}^{2}j$

and $dr=dxi+dyj+dzk$

$\begin{array}{rcl}W& =& \int \left(x^{2}yi-x{y}^{2}j\right).\left(dxi+dyj+dyk\right)\\ W& =& \int x^{2}ydx-x{y}^{2}dy\end{array}$

The equation of line joining points (1,1) and (4,2) is mentioned below.

$\begin{array}{rcl}\frac{y-1}{x-1}& =& \frac{2-1}{4-1}\\ \frac{y-1}{x-1}& =& \frac{1}{3}\\ x& =& 3y-2\\ dx& =& 3dy\end{array}$

Work done given below:

$\begin{array}{rcl}W& =& \underset{1}{\overset{2}{\int }}{\left(3y-2\right)}^{2}y\left(3dy\right)-\left(3y-2\right)y^{2}dy\\ W& =& \underset{1}{\overset{2}{\int }}(27y^3-36y^2+12y-3y^3+2y^2)dy\\ W& =& \underset{1}{\overset{2}{\int }}\left(24y^3-34y^2+12y\right)dy\\ W& =& \frac{86}{3}\end{array}$

Work done along path from   (1,1) and   (4,2)

(i) From (1,1) to (1,0) , we have that x = 1

Then dx = 0

$\begin{array}{rcl}{W}_1& =& \underset{1}{\overset{0}{\int }}{\left(1\right)}^{2}y\left(0\right)-\left(1\right)y^{2}dy\\ W_1& =& \underset{1}{\overset{0}{\int }}-y^{2}dy\\ W_1& =& \frac{1}{3}\end{array}$

(ii) From (1,0) to (4,0) , we have that y = 0

Then dy = 0

$\begin{array}{rcl}{W}_2& =& \underset{1}{\overset{4}{\int }}{\left(x\right)}^2\left(0\right)dx-x{\left(0\right)}^2\left(0\right)\\ W_2& =& \underset{1}{\overset{4}{\int }}0dx\\ W_2& =& 0\end{array}$

(iii) From (4,0) to (4,2) , we have that x = 4

Then dx = 0

$\begin{array}{rcl}{W}_3& =& \underset{0}{\overset{2}{\int }}{\left(4\right)}^{2}y\left(0\right)-4{\left(y\right)}^{2}dy\\ W_3& =& \underset{0}{\overset{2}{\int }}-4y^{2}dy\\ W_3& =& \frac{-32}{3}\end{array}$

Find the total work done along the path.

$\begin{array}{rcl}W& =& W_1+W_2+W_3\\ W& =& \frac{1}{3}+0-\frac{32}{3}\\ W& =& \frac{-31}{3}\end{array}$

Hence the total work done is $\frac{86}{3}$ and work done along the given path is $\frac{-31}{3}$