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2021-01-19

In a truck-loading station at a post office, a small 0.200-kg package is released from rest a point A on a track that is one quarter of a circle with radius 1.60 m.The size of the packageis much less than 1.60m, so the package can be treated at aparticle. It slides down the track and reaches point B with a speed of 4.80 m/s. From point B, it slides on a level surface a distanceof 3.00 m to point C, where is comes to rest.
(a) What is the coefficient of kinetic friction on the horizontal surface?
(b) How much work is done on the package by friction as it slides down the circular arc from A to B?

Elberte

Expert

a.) The kinetic energy of the package at point determines how much work is done by friction.
${\mu }_{k}mgL=\frac{1}{2}m{v}_{B}^{2}$
Then solve for ${\mu }_{k}$
${\mu }_{k}=\frac{{v}_{B}^{2}}{2gL}=\frac{{\left(4.8\right)}^{2}}{2\left(9.8\right)\left(3\right)}=0.392$
b)
${W}_{\text{other}}={K}_{B}-{U}_{A}$
${W}_{\text{other}}=\frac{1}{2}m{v}^{2}-mgh$
$=\frac{1}{2}\left(0.2\right){\left(4.8\right)}^{2}-\left(0.2\right)\left(9.8\right)\left(1.6\right)$
$=-0.832$ J

Jeffrey Jordon

Expert

a) Kinetic friction on horizontal surface is equal to friction force

$\mu Nd\mathrm{cos}\left(\theta \right)=\frac{1}{2}m\left({v}_{2}^{2}-{v}_{1}^{2}\right)$

N=mg

Given

$m=0.200kg;d=3.00m;{v}_{1}=4.80m/s;{v}_{2}=0$

Therefore

$\mu \left(0.200kg\right)×9.8×3\mathrm{cos}{180}^{\circ }=\frac{1}{2}\left(0.200\right)×\left({0}^{2}-{4.80}^{2}\right)$

Step 2

b) Total work= work done by gravity + work done by friction

Total work = kinetic energy

${W}_{T}=\frac{1}{2}m\left({v}_{1}^{2}-{u}^{2}\right)$

${W}_{g}+{W}_{f}=\frac{1}{2}m\left({v}_{1}^{2}-{u}^{2}\right)$

${W}_{f}=\frac{1}{2}m\left({v}_{1}^{2}-{u}^{2}\right)-mgR=\frac{1}{2}0.200\left(\left(4.80{\right)}^{2}-0\right)\right)-\left(0.200\right)\left(9.8\right)\left(1.60\right)$

${W}_{f}=-0.832J$ (work done against moving body , hence -ive sign)

Jeffrey Jordon

Expert

Step 1

Given:

The mass is 2 kg

The package's size is 1.6 m

The speed is given as 4.80 m/s

The distance is 3 m

Step 2

a) At point B, kinetic energy is equal to the friction's work done. The formula is denoted as,

$\frac{1}{2}m{v}_{B}^{2}={\mu }_{k}mg×Bc$

${\mu }_{k}=\frac{{v}_{B}^{2}}{2g×Bc}=\frac{{4.8}^{2}}{2×9.81×3}$

${\mu }_{k}=0.39$

Step 3

b) The sum of kinetic energy at point B and work done by the friction is equal to the Potential energy at point A.

The formula is denoted as,

$mgR={W}_{{F}_{AB}}+\frac{1}{2}m{v}_{B}^{2}$

${W}_{{F}_{AB}}=m\left(gR-\frac{1}{2}{v}_{B}^{2}\right)$

On substituting the values,

${W}_{{F}_{AB}}=0.2\left(9.81×1.60-0.5×{4.8}^{2}\right)$

${W}_{{F}_{AB}}=0.835J$

Step 4

Solution:

a) At the horizontal surface, the coefficient of friction is 0.39

b) From A to B, the friction's work done is 0.835 J

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