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15P

Expert-verifiedFound in: Page 171

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 ${\mathbf{\theta}}{\mathbf{=}}{\mathbf{60}}{\mathbf{.}}{{\mathbf{0}}}^{{\mathbf{\xb0}}}$. 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?**

- Net work done on the trunk by all forces will be 1.5 J.
- The kinetic energy will increase by 1.5 J.

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, ${\mathrm{\theta}}_{2}=\mathrm{\theta}=60.{0}^{\xb0}$

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

**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.**

Free body diagram for the trunk:

** **

** **

Work done by force $\overrightarrow{{\mathrm{F}}_{1}}$ can be calculated as,

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

From the diagram, we can say that the angle between ${\mathrm{F}}_{1}\mathrm{and}\mathrm{d}$ will be ${0}^{\xb0}$, then the work done will be,

${\mathrm{W}}_{1}=5\times 3\times \mathrm{cos}\left({0}^{\xb0}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{W}}_{1}=15\mathrm{J}$

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

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

From the diagram, we can say that the angle between ${\mathrm{F}}_{2}\mathrm{and}\mathrm{d}$ will be ${120}^{\xb0}$, then the work done will be,

${\mathrm{W}}_{2}=9\times 3\times \mathrm{cos}\left({120}^{\xb0}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{W}}_{2}=-13.5\mathrm{J}$

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

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

From the diagram, we can say that the angel between ${\mathrm{F}}_{3}\mathrm{and}\mathrm{d}$ will be ${90}^{\xb0}$, then the work done will be,

${\mathrm{W}}_{3}=3\times 3\times \mathrm{cos}{90}^{\xb0}\phantom{\rule{0ex}{0ex}}{\mathrm{W}}_{3}=0\mathrm{J}$

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 $1.5\mathrm{J}$.

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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