91影视

The given force vector is $F=\left(2xy-3\right)i+x^{2}j$ and path from (1,0) to (0,1)

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.

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

$W=鈭�F.dr$

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.

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

Use the formula $W=鈭�F.dr$

Put $F=\left(2xy-3\right)i+x^{2}j$

and $dr=dxi+dyj+dzk$

$W=鈭�\left(2xy-3\right)dx-x^{2}dy$

The equation of line joining points (1,0) and (0,1) is mentioned below.

$\begin{array}{rcl}\frac{y-1}{x-0}& =& \frac{1-0}{0-1}\\ \frac{y-1}{x}& =& -1\\ y& =& 1-x\\ dy& =& -dx\end{array}$

Solve further.

$\begin{array}{rcl}W& =& \underset{1}{\overset{0}{鈭�}}\left(2x\left(1-x\right)-3\right)dx+x^2\left(-dx\right)\\ W& =& \underset{1}{\overset{0}{鈭�}}\left(2x-2x^2-x^2-3\right)dx\\ W& =& \underset{1}{\overset{0}{鈭�}}\left(2x-3x^2-3\right)dx\\ W& =& 3\end{array}$

From (1,0) to (0,1) on a circular arc ,

Put

$$\begin{gathered} x=r\cos 胃 \\ y=r\sin 胃 \end{gathered}$$

But r = 1

$\begin{array}{rcl}x& =& \cos 胃\\ y& =& \sin 胃\\ dx& =& -\sin 胃d胃\\ dy& =& \cos 胃d胃\end{array}$

$\begin{array}{rcl}W& =& \underset{0}{\overset{\frac{蟺}{2}}{鈭�}}\left(2\cos 胃\sin 胃-3\right)\left(-\sin 胃\right)d胃+\left({\cos }^2胃\right)\cos 胃d胃\\ W& =& \underset{0}{\overset{\frac{蟺}{2}}{鈭�}}\left(-2\cos 胃{\sin }^2胃+3\sin 胃\right)d胃+{\cos }^3胃d胃\\ W& =& \underset{0}{\overset{\frac{蟺}{2}}{鈭�}}\cos 胃\left(-{\sin }^2胃+1\right)d胃+\underset{0}{\overset{\frac{蟺}{2}}{鈭�}}3\sin 胃d胃\end{array}$

Let $u=\sin 胃$

Then $du=\cos 胃d胃$$du=\cos 胃d胃$ .

$\begin{array}{rcl}W& =& \underset{\sin 0}{\overset{\sin \frac{蟺}{2}}{鈭�}}\left(-u^2+1\right)du+\underset{0}{\overset{\frac{蟺}{2}}{鈭�}}3\sin 胃d胃\\ W& =& -\frac{u^3}{3}{\left.u\right|}_{\sin 0}^{\sin \frac{蟺}{2}}-{\left.3\cos 胃\right|}_0^{\frac{蟺}{2}}\\ W& =& 3\end{array}$

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

Then dx = 0

$$\begin{gathered} W_1=\underset{0}{\overset{1}{鈭�}}\left(2\left(1\right)y-3\right)\left(0\right)-1^{2}dy \\ {W}_1=1 \end{gathered}$$

From (1,1) to (0,1) , we have y = 1

Then dy = 0

$$\begin{gathered} W_2=\underset{1}{\overset{0}{鈭�}}\left(2x\left(1\right)-3\right)dx-x^2\left(0\right) \\ {W}_2=2 \end{gathered}$$

Total work done is given below.

$\begin{array}{rcl}W& =& W_1+W_2\\ W& =& 1+2\\ W& =& 3\end{array}$

The work done (a) along straight line W = 3 (b) along circular arc W = 3 (c) along lines parallel to axes W = 3 .