91Ó°ÊÓ

The force acting in the x-direction is $\stackrel{â‡\P Ä}{\mathrm{F}}=\mathrm{Î\pm ³\ae ²â}\hat{\mathrm{i}}$.

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

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,

$\mathbf{W}\mathbf{=}\mathbf{Fs}\mathbf{}\mathbf{³¦´Ç²õθ}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

(a)

The work done can be calculated as follows.

$$\begin{gathered} \mathrm{W}={∫}_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\stackrel{â‡\P Ä}{\mathrm{F}}\mathrm{dx} \\ ={∫}_0^2\left(\mathrm{Î\pm ³\ae ²â}\right)\mathrm{dx} \\ =\left(2.50\mathrm{N}/{\mathrm{m}}^2\right)\left(3\mathrm{m}\right){∫}_0^2\mathrm{xdx} \\ =\left(2.50\mathrm{N}/{\mathrm{m}}^2\right)\left(3\mathrm{m}\right){\left[\frac{{\mathrm{x}}^2}{2}\right]}_0^2 \end{gathered}$$

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 $\mathrm{θ}={90}^{∘}$.

The work done can be calculated as follows,

$$\begin{gathered} \mathrm{W}=\mathrm{Fs³¦´Ç²õθ} \\ =\mathrm{Fscos}\left({90}^{∘}\right) \\ =0\mathrm{J} \end{gathered}$$

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

(c)

The work done can be calculated as follows.

$$\begin{gathered} \mathrm{W}={∫}_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\stackrel{â‡\P Ä}{\mathrm{F}}\mathrm{dx} \\ ={∫}_0^2\left(\mathrm{Î\pm ³\ae ²â}\right)\mathrm{dx} \\ =\left(2.50\mathrm{N}/{\mathrm{m}}^2\right){∫}_0^2\left(1.5\mathrm{x}\right)\mathrm{xdx} \\ =\left(2.50\mathrm{N}/{\mathrm{m}}^2\right)\left(1.5\right){\left[\frac{{\mathrm{x}}^3}{3}\right]}_0^2 \end{gathered}$$

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.