An object has several forces acting on it. One of these forces is $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}\mathbf{=}\mathbf{\alpha xy}\hat{\mathbf{i}}$, a force in the x-direction whose magnitude depends on the position of the object, with $\mathbf{\alpha }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{N}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. Calculate the work done on the object by this force for the following displacements of the object: (a) The object starts at the point$\left(x=0,y=3.00m\right)$ and moves parallel to the x-axis to the point role="math" localid="1668072687742" $(x=2.00,y=3.00m)$. (b) The object starts at the point$\left(x=2.00,y=0\right)$ and moves in the y-direction to the point, $\left(x=2.00,y=3.00m\right)$. (c) The object starts at the origin and moves on the line$\mathbf{y}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{x}$ to the point $\left(x=2.00,y=3.00m\right)$.

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

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

The object starts at the point$\left(\mathrm{x},\mathrm{y}\right)=\left(0,3\right)$ and the object reaches the point $\left(2,3\right)$.

The expression of work done is given by,

$\mathbf{W}\mathbf{=}\mathbf{Fs}\mathbf{}\mathbf{cos\theta }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

(a)

The work done can be calculated as follows.

$$\begin{gathered} \mathrm{W}={\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\stackrel{\rightharpoonup }{\mathrm{F}}\mathrm{dx} \\ ={\int }_0^2\left(\mathrm{\alpha xy}\right)\mathrm{dx} \\ =\left(2.50\mathrm{N}/{\mathrm{m}}^2\right)\left(3\mathrm{m}\right){\int }_0^2\mathrm{xdx} \\ =\left(2.50\mathrm{N}/{\mathrm{m}}^2\right)\left(3\mathrm{m}\right){\left[\frac{{\mathrm{x}}^2}{2}\right]}_0^2 \end{gathered}$$

Simplify further,

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

Therefore, the required work done is 15.0 J.

(b)

The object is moving in the perpendicular direction, so $\theta ={90}^{\circ }$.

The work done can be calculated as follows,

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

Therefore, the required work done is $0\mathrm{J}$.

(c)

The work done can be calculated as follows.

$$\begin{gathered} \mathrm{W}={\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\stackrel{\rightharpoonup }{\mathrm{F}}\mathrm{dx} \\ ={\int }_0^2\left(\mathrm{\alpha xy}\right)\mathrm{dx} \\ =\left(2.50\mathrm{N}/{\mathrm{m}}^2\right){\int }_0^2\left(1.5\mathrm{x}\right)\mathrm{xdx} \\ =\left(2.50\mathrm{N}/{\mathrm{m}}^2\right)\left(1.5\right){\left[\frac{{\mathrm{x}}^3}{3}\right]}_0^2 \end{gathered}$$

Simplify further.

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

Therefore, the required work done is 10.0 J.