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Q101P

Expert-verifiedFound in: Page 148

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Question: Playing near a road construction site, a child falls over a barrier and down onto a dirt slope that is angled downward at **** to the horizontal. As the child slides down the slope, he has an acceleration that has a magnitude of 0.50 m/s^{2}**

**Answer**

The coefficient of friction between the child and slope is 0.76.

Angle = 35^{0}

Acceleration of the child = 0.50 m/s^{2}

**The frictional force is given by the product of the coefficient of friction and the normal reaction. Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass. Using these concepts, the problem can be solved.**

In case of “downhill” (positive), then Newton’s law gives

$mgsin\theta -{f}_{k}=ma$

for the sliding child,

${f}_{k}={\mu}_{k}{F}_{N}={\mu}_{k}mg\phantom{\rule{0ex}{0ex}}a=g\left(\text{sin}\theta -{\mu}_{k}\text{cos}\theta \right)\phantom{\rule{0ex}{0ex}}=-0.50{\text{m/s}}^{\text{2}}\phantom{\rule{0ex}{0ex}}g\left(\text{sin}\theta -{\mu}_{k}\text{cos}\theta \right)=-0.50{\text{m/s}}^{\text{2}}\phantom{\rule{0ex}{0ex}}$

Rearranging for the${\mu}_{k}$, we get

${\mu}_{k}=\frac{1}{\text{cos}\theta}\left(\frac{0.50{\text{m/s}}^{\text{2}}}{g}+\text{sin}\theta \right)$

Substituting the value of ${\mu}_{k}$

With localid="1661258346840" $\theta ={35}^{0}$ , we solve for the coefficient .

${\mu}_{k}=\frac{1}{\text{cos}35}\left(\frac{0.50{\text{m/s}}^{\text{2}}}{9.8}+\text{sin}35\right)\phantom{\rule{0ex}{0ex}}=0.76\phantom{\rule{0ex}{0ex}}$

Therefore, the coefficient of friction between the child and slope is 0.76.

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