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Q15P

Expert-verifiedFound in: Page 208

Book edition
4th edition

Author(s)
Ruth W. Chabay, Bruce A. Sherwood

Pages
1135 pages

ISBN
9781118875865

**In Figure 5.72 ${{\mathbf{m}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{12}}{\mathbf{\hspace{0.33em}}}{\mathbf{kg}}$and ${{\mathbf{m}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{5}}{\mathbf{\hspace{0.33em}}}{\mathbf{kg}}$ ****. The kinetic coefficient of friction between ${{\mathbf{m}}}_{{\mathbf{1}}}$ **** and the floor is ${\mathbf{0}}{\mathbf{.}}{\mathbf{3}}$**** and that between ${{\mathbf{m}}}_{{\mathbf{2}}}$**** and the floor is ${\mathbf{0}}{\mathbf{.}}{\mathbf{5}}$****. You push with a force of magnitude ${\mathbf{F}}{\mathbf{=}}{\mathbf{110}}{\mathbf{\hspace{0.33em}}}{\mathbf{N}}$ ****. (a) What is the acceleration of the center of mass? (b) What is the magnitude of the force that ${{\mathbf{m}}}_{{\mathbf{1}}}$**** exerts on ${{\mathbf{m}}}_{{\mathbf{2}}}$****?**

(a) The acceleration of the center of mass is $2.95\mathrm{m}/{\mathrm{s}}^{2}$.

(b) The magnitude of the force that ${m}_{1}$exerts on ${m}_{2}$ is $39.27\hspace{0.33em}\mathrm{N}$.

The given data can be listed below as:

- The value of the first mass is,${\mathrm{m}}_{1}=12\hspace{0.33em}\mathrm{kg}$ .
- The value of the second mass is,${\mathrm{m}}_{2}=5\hspace{0.33em}\mathrm{kg}$ .
- The coefficient of friction between the floor and the first mass is ${\mu}_{1}=0.3$.
- The coefficient of friction between the floor and the second mass is ${\mu}_{2}=0.5$ .
- The force required to push the masses are $\mathrm{F}=110\hspace{0.33em}\mathrm{N}$ .

**The coefficient of friction is described as the ratio between the friction and the normal force. However, the coefficient of friction mainly depends on surface roughness and nature of the material used.**

The equation of the net force is expressed as:

${\mathrm{F}}_{\mathrm{net}}=\mathrm{F}-({\mathrm{\mu}}_{1}{\mathrm{m}}_{1}\mathrm{g}+{\mathrm{\mu}}_{2}{\mathrm{m}}_{2}\mathrm{g})$

Here,${F}_{net}$ is the net force, F is the force required to push the masses, ${\mathrm{\mu}}_{1}$is the coefficient of friction between the floor and the first mass,${m}_{1}$ is the mass of the first mass, g is the acceleration due to gravity, ${\mu}_{2}$ is the coefficient of friction between the floor and the second mass and ${m}_{2}$ is the mass of the second mass.

Substitute the values in the above equation.

${\mathrm{F}}_{\mathrm{net}}=110\mathrm{N}-\left(0.3\times 12\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^{2}+0.5\times 5\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^{2}\right)\phantom{\rule{0ex}{0ex}}=110\mathrm{N}-\left(35.28\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}+24.5\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}\right)\phantom{\rule{0ex}{0ex}}=110\mathrm{N}-\left(59.78\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}\times \left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}}\right)\right)\phantom{\rule{0ex}{0ex}}=50.22\mathrm{N}$

The equation of the acceleration of the center of mass is expressed as:

$\mathrm{a}=\frac{{\mathrm{F}}_{\mathrm{net}}}{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}}$

Here, a is the acceleration of the center of mass,${F}_{net}$ is the net force, ${m}_{1}$ is the mass of the first mass and ${m}_{2}$is the mass of the second mass.

Substitute the values in the above equation.

$\mathrm{a}=\frac{50.22\mathrm{N}}{12\mathrm{kg}+5\mathrm{kg}}\phantom{\rule{0ex}{0ex}}=\frac{50.22\mathrm{N}}{17\mathrm{kg}}\phantom{\rule{0ex}{0ex}}=2.95\mathrm{N}/\mathrm{kg}\phantom{\rule{0ex}{0ex}}=2.95\mathrm{N}/\mathrm{kg}\times \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}/1\mathrm{N}\phantom{\rule{0ex}{0ex}}=2.95\mathrm{m}/{\mathrm{s}}^{2}$

Thus, the acceleration of the center of mass is $2.95m/{s}^{2}$.

** **The equation of the magnitude of the force that the first mass exerts on the second mass is expressed as:

${\mathrm{F}}_{1}={\mathrm{m}}_{2}\mathrm{a}+{\mathrm{\mu}}_{2}{\mathrm{m}}_{2}\mathrm{g}$

Here, ${F}_{1}$ is the magnitude of the force that the first mass exerts on the second mass, a is the acceleration of the center of mass, g is the acceleration due to gravity, ${\mu}_{2}$ is the coefficient of friction between the floor and the second mass and ${m}_{2}$ is the mass of the second mass.

Substitute the values in the above equation.

${\mathrm{F}}_{1}=\left(5\mathrm{kg}\right)\left(2.95\mathrm{m}/{\mathrm{s}}^{2}\right)+\left(0.5\right)\left(5\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^{2}\right)\phantom{\rule{0ex}{0ex}}=14.77\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}+24.5\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}\phantom{\rule{0ex}{0ex}}=39.27\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}\times \left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=39.27\mathrm{N}$

Thus, the magnitude of the force that ${m}_{1}$ exerts on ${m}_{2}$ is$39.27\hspace{0.33em}\mathrm{N}$ .

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