Question

How much work is done by a force $\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{(}\mathbf{2}\mathit{x}\mathbf{N}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{3}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{j}}$,with x in meters, that moves a particle from a position $\overrightarrow{{\mathbf{r}}_{\mathbf{i}}}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{}\mathbf{m}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{3}\mathbf{}\mathbf{m}\mathbf{)}\hat{\mathit{j}}$to a position $\overrightarrow{{\mathbf{r}}_{\mathbf{f}}}\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{4}\mathbf{}\mathbf{m}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{3}\mathbf{}\mathbf{m}\mathbf{)}\hat{\mathbf{j}}$?

Short answer

Work done by the applied force is $-6\mathrm{J}$

Step-by-step solution

Given information

It is given that,

$$\begin{gathered} F_{app}=(2x\mathrm{N})\hat{\mathrm{i}}+(3\mathrm{N})\hat{\mathrm{j}} \\ \overrightarrow{r_i}=(2\mathrm{m})\hat{\mathrm{i}}+(3\mathrm{m})\hat{\mathrm{j}} \\ \overrightarrow{r_f}=-(4\mathrm{m})\hat{\mathrm{i}}-(3\mathrm{m})\hat{\mathrm{j}} \end{gathered}$$

Determining the concept

This problem deals with the variable forces. When the direction and the amount of force vary throughout the motion, variable forces occur. To solve this problem, usetheconcept of variable forces. Work done due to variable forces can be found by using integration.

Formula:

The work done is given by,

$W=\int F_x.dx+\int F_y.dy$

Where,

$F_{x,}{F}_y$Are the forces in x and y direction, dx and dy are small displacements.

Determining the work done by the applied force (Wapp).

Using the above formula, writethetotal work done as,

$W={\int }_{xi}^{xf}{F}_{x}dx+{\int }_{yi}^{yf}{F}_{y}dy$

Substituting the value of forces and from the given data,

$$\begin{gathered} W={\int }_2^{-4}2\times dx+{\int }_3^{-3}3dy \\ W={\left[x^2\right]}_2^{-4}+{\left[3y\right]}_3^{-3} \\ W=16-4-9-9 \\ W=-6\mathrm{J} \end{gathered}$$

Work done by the applied force is $-6\mathrm{J}$