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Expert-verified Found in: Page 173 ### Fundamentals Of Physics

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 $\mathrm{W}=25\mathrm{J}$.

See the step by step solution

## Step 1: Given

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

## Step 2: Understanding the concept

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

Formula:

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

## Step 3: Calculate the net work done

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.

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

Therefore, the net work done is 25J ### Want to see more solutions like these? 