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Expert-verified Found in: Page 171 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Figure 7-28 shows three forces applied to a trunk that moves leftward by 3.00 m over a frictionless floor. The force magnitudes are ${{\mathbf{F}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{5}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{N}}$, ${{\mathbf{F}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{9}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{N}}$, and ${{\mathbf{F}}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{3}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{N}}$, and the indicated angle is ${\mathbf{\theta }}{\mathbf{=}}{\mathbf{60}}{\mathbf{.}}{{\mathbf{0}}}^{{\mathbf{°}}}$. During the displacement, (a) what is the net work done on the trunk by the three forces and (b) does the kinetic energy of the trunk increase or decrease? 1. Net work done on the trunk by all forces will be 1.5 J.
2. The kinetic energy will increase by 1.5 J.
See the step by step solution

## Step 1: Given data

The magnitude of forces are, ${\mathrm{F}}_{1}=5.0\mathrm{N}$, ${\mathrm{F}}_{2}=9.0\mathrm{N}$, ${\mathrm{F}}_{3}=3.0\mathrm{N}$.

The given angle is, ${\mathrm{\theta }}_{2}=\mathrm{\theta }=60.{0}^{°}$

The traveled distance is, $\mathrm{d}=3.0\mathrm{m}$ to the left.

## Step 2: Understanding the concept

As the trunk is moving to the left, we can calculate the work done by each force separately. We can find the sum of the individual work done due to each force to calculate the net work done.

## Step 3: (a) Calculate the net work done on the trunk by the three forces

Free body diagram for the trunk: Work done by force $\stackrel{\to }{{\mathrm{F}}_{1}}$ can be calculated as,

$\mathrm{W}=\stackrel{\to }{{\mathrm{F}}_{1}}.\stackrel{\to }{\mathrm{d}}\phantom{\rule{0ex}{0ex}}{\mathrm{W}}_{1}={\mathrm{F}}_{1}\mathrm{dcos\varphi }$

From the diagram, we can say that the angle between ${\mathrm{F}}_{1}\mathrm{and}\mathrm{d}$ will be ${0}^{°}$, then the work done will be,

${\mathrm{W}}_{1}=5×3×\mathrm{cos}\left({0}^{°}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{W}}_{1}=15\mathrm{J}$

Work done by force $\stackrel{\to }{{\mathrm{F}}_{2}}$ can be calculated as,

${\mathrm{W}}_{2}=\stackrel{\to }{{\mathrm{F}}_{2}}.\stackrel{\to }{\mathrm{d}}\phantom{\rule{0ex}{0ex}}{\mathrm{W}}_{2}={\mathrm{F}}_{2}\mathrm{dcos\varphi }$

From the diagram, we can say that the angle between ${\mathrm{F}}_{2}\mathrm{and}\mathrm{d}$ will be ${120}^{°}$, then the work done will be,

${\mathrm{W}}_{2}=9×3×\mathrm{cos}\left({120}^{°}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{W}}_{2}=-13.5\mathrm{J}$

Work done by force $\stackrel{\to }{{\mathrm{F}}_{3}}$ can be calculated as,

$\mathrm{W}=\stackrel{\to }{{\mathrm{F}}_{3}}.\stackrel{\to }{\mathrm{d}}\phantom{\rule{0ex}{0ex}}\mathrm{W}={\mathrm{F}}_{3}\mathrm{dcos\varphi }$

From the diagram, we can say that the angel between ${\mathrm{F}}_{3}\mathrm{and}\mathrm{d}$ will be ${90}^{°}$, then the work done will be,

${\mathrm{W}}_{3}=3×3×\mathrm{cos}{90}^{°}\phantom{\rule{0ex}{0ex}}{\mathrm{W}}_{3}=0\mathrm{J}$

Total work done can be calculated as,

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

Substitute the values in the above expression, and we get,

${\mathrm{W}}_{\mathrm{Net}}=15\mathrm{J}-13.5\mathrm{J}+0\mathrm{J}\phantom{\rule{0ex}{0ex}}{\mathrm{W}}_{\mathrm{Net}}=1.5\mathrm{J}$

Thus, the net work done on the trunk by all forces will be $1.5\mathrm{J}$.

## Step 4: (b) Find out if the kinetic energy of the trunk increases or decreases

According to work–energy theorem, if there are no forces acting other than these three forces, during the displacement, the kinetic energy of the trunk will be equal to the net work done.

The kinetic energy will increase, and it will increase by 1.5 J.

Thus, the kinetic energy will increase by 1.5 J.

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