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36P

Expert-verifiedFound in: Page 173

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A 5.0 kg** **block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown in Fig.7-39****. The scale of the figure’s vertical axis is set by ${{\mathbf{F}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{10}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{N}}$****. How much work is done by the force as the block moves from the origin to** ${\mathbf{x}}{\mathbf{=}}{\mathbf{8}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{cm}}$**?**

The net work done by the force using a graph $\mathrm{W}=25\mathrm{J}$.

- The mass of block is, $\mathrm{m}=5.0\mathrm{kg}$
- The scale for vertical axis is, ${\mathrm{F}}_{\mathrm{s}}=10.0\mathrm{N}$

**We can use the concept that the work done is equal to the area under the curve for a graph.**

**Formula:**

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

We know that the total area will be the net work done by the force.

So,

$\mathrm{W}={\mathrm{W}}_{1}+{\mathrm{W}}_{2}+{\mathrm{W}}_{3}+{\mathrm{W}}_{4}$

Substitute all the value in the above equation.

$\mathrm{W}=10\mathrm{N}\times 2\mathrm{m}+\frac{1}{2}\times 10\mathrm{N}\times 2\mathrm{m}+0+\left(-\frac{1}{2}\times 2\mathrm{N}\times 5\mathrm{m}\right)\phantom{\rule{0ex}{0ex}}\mathrm{W}=25\mathrm{J}$

Therefore, the net work done is 25J

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