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Expert-verified Found in: Page 167 ### Matter & Interactions

Book edition 4th edition
Author(s) Ruth W. Chabay, Bruce A. Sherwood
Pages 1135 pages
ISBN 9781118875865 # A ${\mathbf{5}}{\mathbf{}}{\mathbit{k}}{\mathbit{g}}$ box with initial speed ${\mathbf{4}}{\mathbit{m}}{\mathbf{/}}{\mathbit{s}}$slides across the floor and comes to a stop after ${\mathbf{0}}{\mathbf{.}}{\mathbf{7}}{\mathbit{s}}$ (a) what is the coefficient of kinetic friction? (b) How far does the box move? (c) You put a ${\mathbf{3}}{\mathbit{k}}{\mathbit{g}}$ block in the box, so the total mass is now ${\mathbf{8}}{\mathbit{k}}{\mathbit{g}}$, and you launch this heavier box with an initial speed of ${\mathbf{4}}{\mathbit{m}}{\mathbf{/}}{\mathbit{s}}$. How long does it take to stop?

1. The coefficient of kinetic friction is $1.17$

2. The box moves $0.7m$

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

See the step by step solution

## Step 1: Identification of the given data

The mass of a box is 5 kg.

The initial speed is 4m/s.

Time taken to stop the box is 0.7s.

## Step 2: Definition of frictional force and the coefficient of kinetic friction

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.

## Step 3: (a) Determination of the coefficient of kinetic friction

Coefficient of kinetic friction,

${\mathbit{\mu }}{\mathbf{=}}\frac{\mathbf{F}}{\mathbf{N}}$ ..(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),

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

Substitute $“a”$ value in Equation (2),

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

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),

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

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" ${\mathbf{1}}{\mathbf{.}}{\mathbf{17}}$

## Step 4: (b) Determination of the distance moved by the box

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

Where, d =distance

From the Equation (5),

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

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

## Step 5: (c) Determination of the distance moved by the box

Time taken to stop the box,

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

Hence, the Time taken to stop the box is ${\mathbf{0}}{\mathbf{.}}{\mathbf{56}}{\mathbit{s}}$ ### Want to see more solutions like these? 