An object has several forces acting on it. One of these forces is $\overrightarrow{\mathbf{F}}\mathbf{=}\mathit{\alpha }\mathit{x}\mathit{y}\hat{\mathbf{i}}$, a force in the x-direction whose magnitude depends on the position of the object, with $\mathbf{\alpha }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{}\mathit{N}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$. Calculate the work done on the object by this force for the following displacements of the object: (a) The object starts at the point (x = 2.00, y = 3.00 m) and moves parallel to the x-axis to the point (x = 0, y = 3.00 m). (b) The object starts at the point and moves in the y-direction to the point, (x = 2.00, y = 3.00 m) . (c) The object starts at the origin and moves on the line to the point (x = 2.00, y = 3.00 m) .

The force acting in the x-direction is $\overrightarrow{\mathrm{F}}=\alpha xy\hat{i}$.

The value of $\alpha =2.50N/m^2$.

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,

$\mathit{W}\mathbf{=}\mathit{F}\mathbf{}\mathit{s}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }$…………………(1)

The work done can be calculated as follows.

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

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 $\theta ={90}^o$.

The work done can be calculated as follows,

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

Therefore, the required work done is 0 J.

The work done can be calculated as follows.

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

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.