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Q42P

Expert-verifiedFound in: Page 167

Book edition
4th edition

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

Pages
1135 pages

ISBN
9781118875865

**A ${\mathbf{5}}{\mathbf{}}{\mathit{k}}{\mathit{g}}$ box with initial speed ${\mathbf{4}}{\mathit{m}}{\mathbf{/}}{\mathit{s}}$slides across the floor and comes to a stop after ${\mathbf{0}}{\mathbf{.}}{\mathbf{7}}{\mathit{s}}$ (a) what is the coefficient of kinetic friction? (b) How far does the box move? (c) You put a ${\mathbf{3}}{\mathit{k}}{\mathit{g}}$ block in the box, so the total mass is now ${\mathbf{8}}{\mathit{k}}{\mathit{g}}$, and you launch this heavier box with an initial speed of ${\mathbf{4}}{\mathit{m}}{\mathbf{/}}{\mathit{s}}$. How long does it take to stop?**

The coefficient of kinetic friction is $1.17$

The box moves $0.7m$

The time taken to stop a box of $8kg$ is $0.56s$

The mass of a box is **5 kg**.

The initial speed is **4m/s.**

Time taken to stop the box is **0.7s.**

Frictional force is defined as **the force required resisting the motion of a box when one object’s surface contacts with other ones object.**

The coefficient of kinetic friction is defined as **the ratio of kinetic friction force to normal force.**

Coefficient of kinetic friction,

${\mathit{\mu}}{\mathbf{=}}\frac{\mathbf{F}}{\mathbf{N}}$ ..(1)

Where, F = Frictional Force

N = Normal Force

To find Frictional Force,

$F=ma$ …(2)

Where, m = mass of a box

a = acceleration

To find acceleration a,

${v}^{2}={u}^{2}+2as$ ..(3)

Where, $v=0$

$u=4m/s$

$s=0.7s$

Substitute these values in Equation (3),

$0={\left(4\right)}^{2}+\left(2\times a\times 0.7\right)\phantom{\rule{0ex}{0ex}}0=16+1.4a\phantom{\rule{0ex}{0ex}}1.4a=-16\phantom{\rule{0ex}{0ex}}a=11.43m/{s}^{2}$

Substitute $\u201ca\u201d$ value in Equation (2),

$F=ma\phantom{\rule{0ex}{0ex}}F=5\times 11.43\phantom{\rule{0ex}{0ex}}F=57.14N$

Frictional Force is $57.14N$

To find Normal Force $N$,

$N=mg$ …(4)

Where,

m= mass

$g=9.8m/{s}^{2}$

Substitute these values in Equation (4),

$N=mg\phantom{\rule{0ex}{0ex}}N=5\times 9.8\phantom{\rule{0ex}{0ex}}N=49N$

From Equation (1),

Coefficient of kinetic friction,

$\mu =\frac{F}{N}\phantom{\rule{0ex}{0ex}}=\frac{57.14}{49}\phantom{\rule{0ex}{0ex}}=1.17$

**Hence, the Coefficient of kinetic friction is role="math" ${\mathbf{1}}{\mathbf{.}}{\mathbf{17}}$**

${\mathit{F}}{\mathbf{\times}}{\mathit{d}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{m}}{{\mathit{v}}}^{{\mathbf{2}}}$ …(5)

Where, d =distance

From the Equation (5),

$F\times d=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}d=\frac{1}{2}\frac{m{v}^{2}}{F}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times \frac{\left(5\times {4}^{2}\right)}{57.14}\phantom{\rule{0ex}{0ex}}=0.7m$

**Hence, the box moves ${\mathbf{0}}{\mathbf{.}}{\mathbf{7}}{\mathit{m}}$.**

Time taken to stop the box,

${\mathit{F}}{\mathbf{=}}\frac{\mathbf{m}\mathbf{(}\mathbf{v}\mathbf{-}\mathbf{u}\mathbf{)}}{\mathbf{t}}\phantom{\rule{0ex}{0ex}}\mathit{t}\mathbf{=}\frac{\mathbf{m}\mathbf{(}\mathbf{v}\mathbf{-}\mathbf{u}\mathbf{)}}{\mathbf{F}}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{=}\frac{\mathbf{8}\mathbf{(}\mathbf{0}\mathbf{-}\mathbf{4}\mathbf{)}}{\mathbf{57}\mathbf{.}\mathbf{14}}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{56}\mathit{s}$

**Hence, the Time taken to stop the box is ${\mathbf{0}}{\mathbf{.}}{\mathbf{56}}{\mathit{s}}$**

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