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35P

Expert-verifiedFound in: Page 1

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**The force on a particle is directed along an x axis and given by **${\mathbf{F}}{\mathbf{=}}{{\mathbf{F}}}_{{\mathbf{0}}}{\left(\frac{x}{{x}_{0}}-1\right)}$**. Find the work done by the force in moving the particle from ${\mathbf{x}}{\mathbf{=}}{\mathbf{0}}{\mathbf{}}{\mathbf{to}}{\mathbf{}}{\mathbf{x}}{\mathbf{=}}{{\mathbf{x}}}_{{\mathbf{0}}}$** **by (a) plotting ${\mathbf{F}}{\left(x\right)}$** **and measuring the work from the graph and (b) integrating **${\mathbf{F}}{\left(x\right)}$**.**

- The net work done by the given force using graph is 0
- The net work done by the given force using integration is 0

The given force function is,

$\mathrm{F}={\mathrm{F}}_{0}\left(\frac{\mathrm{x}}{{\mathrm{x}}_{0}}-1\right)$

The particle moves from $\mathrm{x}=0\mathrm{to}\mathrm{x}=2{\mathrm{x}}_{0}$

**We can use the equation of work for ****the ****integration method. For the graph method, the work done is equal to the area under the curve in the graph.**

**Formula:**

${\mathbf{W}}{\mathbf{=}}{{\mathbf{\int}}}_{{\mathbf{x}}_{\mathbf{i}}}^{{\mathbf{x}}_{\mathbf{f}}}{\mathbf{F}}{\left(x\right)}{\mathbf{d}}{\mathbf{x}}\phantom{\rule{0ex}{0ex}}{\mathbf{A}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\times}}{\mathbf{base}}{\mathbf{\times}}{\mathbf{height}}$

The given function of force is,

$\mathrm{F}={\mathrm{F}}_{0}\left(\frac{\mathrm{x}}{{\mathrm{x}}_{0}}-1\right)$

So,

For x=0 | F = -F |

For x=x | F = 0 |

For x=x | F = F |

We can plot the graph of F(x) using these points as

The work done, from the graph, is simply the area under the curve.

In the above graph,we see that the area under the curve from $\mathrm{x}=0\mathrm{to}\mathrm{x}=2{\mathrm{x}}_{0}$ is equal to the area under the curve for $\mathrm{x}=0\mathrm{to}\mathrm{x}=2{\mathrm{x}}_{0}$ and both are opposite to each other

So, the work done from the graph is $\mathrm{W}=0\mathrm{J}$

Using the integration method, we have

$\mathrm{W}={\int}_{{\mathrm{x}}_{\mathrm{i}}}^{{\mathrm{x}}_{\mathrm{f}}}\mathrm{F}\left(\mathrm{x}\right)d\mathrm{x}\phantom{\rule{0ex}{0ex}}\mathrm{W}={\int}_{0}^{2{\mathrm{x}}_{0}}{\mathrm{F}}_{0}\left(\frac{\mathrm{x}}{{\mathrm{x}}_{0}}-1\right)d\mathrm{x}\phantom{\rule{0ex}{0ex}}\mathrm{W}={\mathrm{F}}_{0}{\left(\frac{{\mathrm{x}}^{2}}{2{\mathrm{x}}_{0}}-\mathrm{x}\right)}_{0}^{2{\mathrm{x}}_{0}}\phantom{\rule{0ex}{0ex}}\mathrm{W}=2{\mathrm{x}}_{0}-2{\mathrm{x}}_{0}\phantom{\rule{0ex}{0ex}}\mathrm{W}=0\mathrm{J}$

Therefore, work done is 0 J.

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