Suggested languages for you:

Americas

Europe

Q61P

Expert-verifiedFound in: Page 175

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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 $\overrightarrow{{\mathbf{r}}_{\mathbf{f}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{(}}{\mathbf{4}}{\mathbf{}}{\mathbf{m}}\mathbf{)}\hat{\mathbf{i}}{\mathbf{+}}{\mathbf{(}}{\mathbf{3}}{\mathbf{}}{\mathbf{m}}{\mathbf{)}}\hat{\mathbf{j}}$****?**

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

It is given that,

${F}_{app}=(2x\mathrm{N})\hat{\mathrm{i}}+(3\mathrm{N})\hat{\mathrm{j}}\phantom{\rule{0ex}{0ex}}\overrightarrow{{r}_{i}}=(2\mathrm{m})\hat{\mathrm{i}}+(3\mathrm{m})\hat{\mathrm{j}}\phantom{\rule{0ex}{0ex}}\overrightarrow{{r}_{f}}=-(4\mathrm{m})\hat{\mathrm{i}}-(3\mathrm{m})\hat{\mathrm{j}}$

**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, use ****the ****concept 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.

Using the above formula, write the total 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,

$W={\int}_{2}^{-4}2\times dx+{\int}_{3}^{-3}3dy\phantom{\rule{0ex}{0ex}}W={\left[{x}^{2}\right]}_{2}^{-4}+{\left[3y\right]}_{3}^{-3}\phantom{\rule{0ex}{0ex}}W=16-4-9-9\phantom{\rule{0ex}{0ex}}W=-6\mathrm{J}$

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

94% of StudySmarter users get better grades.

Sign up for free