91Ó°ÊÓ

The force acting in the x-direction is $\stackrel{→}{\mathrm{F}}=\mathrm{Î\pm }xy\hat{i}$.

The value of $\mathrm{Î\pm }=2.50N/m^2$.

The object starts at the point (x,y ) = (0,3)and the object reaches the point (2,3).

The expression of work done is given by,

$\mathit{W}\mathbf{=}\mathit{F}\mathbf{}\mathit{s}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{θ}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(1)

The work done can be calculated as follows.

$\begin{array}{r}W={∫}_{x_1}^{x_2} \stackrel{→}{F}dx\\ ={∫}_0^2 \left(\mathrm{Î\pm }xy\right)dx\\ =\left(2.50\mathrm{N}/{\mathrm{m}}^2\right)\left(3\mathrm{m}\right){∫}_0^2 xdx\\ =\left(2.50\mathrm{N}/{\mathrm{m}}^2\right)\left(3\mathrm{m}\right){\left[\frac{{\mathrm{x}}^2}{2}\right]}_0^2\end{array}$

Simplify further,

$$\begin{gathered} W=\left(2.50N/m^2\right)\left(3m\right)\left(2-0\right) \\ =15.0J \end{gathered}$$

Therefore, the required work done is 15.0 J.

The object is moving in the perpendicular direction, so $\mathrm{θ}={90}^o$.

The work done can be calculated as follows,

$\begin{array}{r}\mathrm{W}=\mathrm{Fscos}\mathrm{θ}\\ =\mathrm{Fscos}\left({90}^{∘}\right)\\ =0\mathrm{J}\end{array}$

Therefore, the required work done is 0 J.

The work done can be calculated as follows.

$\begin{array}{r}W={∫}_{x_1}^{x_2} \stackrel{→}{F}dx\\ ={∫}_0^2 \left(\mathrm{Î\pm }xy\right)dx\\ =\left(2.50\mathrm{N}/{\mathrm{m}}^2\right){∫}_0^2 \left(1.5x\right)xdx\\ =\left(2.50\mathrm{N}/{\mathrm{m}}^2\right)\left(1.5\right){\left[\frac{x^3}{3}\right]}_0^2\end{array}$

Simplify further.

$$\begin{gathered} W=\left(2.50N/m^2\right)\left(1.5\right)\left(\frac{8}{3}-0\right) \\ =10.0J \end{gathered}$$

Therefore, the required work done is 10.0 J.