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Expert-verified Found in: Page 175 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # How much work is done by a force $\stackrel{\mathbf{\to }}{\mathbf{F}}{\mathbf{=}}{\mathbf{\left(}}{\mathbf{2}}{\mathbit{x}}{\mathbf{N}}\mathbf{\right)}\stackrel{\mathbf{^}}{\mathbf{i}}{\mathbf{+}}{\mathbf{\left(}}{\mathbf{3}}{\mathbf{}}{\mathbf{N}}{\mathbf{\right)}}\stackrel{\mathbf{^}}{\mathbf{j}}$, with x in meters, that moves a particle from a position $\stackrel{\mathbf{\to }}{{\mathbf{r}}_{\mathbf{i}}}{\mathbf{=}}{\mathbf{\left(}}{\mathbf{2}}{\mathbf{}}{\mathbf{m}}\mathbf{\right)}\stackrel{\mathbf{^}}{\mathbf{i}}{\mathbf{+}}{\mathbf{\left(}}{\mathbf{3}}{\mathbf{}}{\mathbf{m}}{\mathbf{\right)}}\stackrel{\mathbit{^}}{\mathbit{j}}$to a position $\stackrel{\mathbf{\to }}{{\mathbf{r}}_{\mathbf{f}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{\left(}}{\mathbf{4}}{\mathbf{}}{\mathbf{m}}\mathbf{\right)}\stackrel{\mathbf{^}}{\mathbf{i}}{\mathbf{+}}{\mathbf{\left(}}{\mathbf{3}}{\mathbf{}}{\mathbf{m}}{\mathbf{\right)}}\stackrel{\mathbf{^}}{\mathbf{j}}$?

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

See the step by step solution

## Step 1: Given information

It is given that,

${F}_{app}=\left(2x\mathrm{N}\right)\stackrel{^}{\mathrm{i}}+\left(3\mathrm{N}\right)\stackrel{^}{\mathrm{j}}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{{r}_{i}}=\left(2\mathrm{m}\right)\stackrel{^}{\mathrm{i}}+\left(3\mathrm{m}\right)\stackrel{^}{\mathrm{j}}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{{r}_{f}}=-\left(4\mathrm{m}\right)\stackrel{^}{\mathrm{i}}-\left(3\mathrm{m}\right)\stackrel{^}{\mathrm{j}}$

## Step 2: 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, 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.

## Step 3: Determining the work done by the applied force (Wapp).

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,

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

Work done by the applied force is $-6\mathrm{J}$ ### Want to see more solutions like these? 