Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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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/s² and that is directed up the slope. What is the coefficient of kinetic friction between the child and the slope?**

Answer

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

Angle = 35⁰

Acceleration of the child = 0.50 m/s²

Step 2: Understanding the concept

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.

Step 3: Calculate the coefficient of kinetic friction between the child and the slope

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

$m g s i n θ - f_{k} = m a$

for the sliding child,

$f_{k} = μ_{k} F_{N} = μ_{k} m g a = g (sin θ - μ_{k} cos θ) = - 0.50 {m/s}^{2} g (sin θ - μ_{k} cos θ) = - 0.50 {m/s}^{2}$

Rearranging for the $μ_{k}$ , we get

$μ_{k} = \frac{1}{cos θ} (\frac{0.50 {m/s}^{2}}{g} + sin θ)$

Substituting the value of $μ_{k}$

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

$μ_{k} = \frac{1}{cos 35} (\frac{0.50 {m/s}^{2}}{9.8} + sin 35) = 0.76$

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

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