Tag Archives: Application of Newton’s Laws

Solution Guides to College Physics by Openstax Chapter 5 Banner

Chapter 5: Further Applications of Newton’s Laws: Friction, Drag, and Elasticity

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Friction

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

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Drag Forces

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

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Elasticity: Stress and Strain

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

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Solution Guides to College Physics by Openstax Chapter 4 Banner

Chapter 4: Dynamics: Force and Newton’s Laws of Motion

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Newton’s Second Law of Motion: Concept of a System

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

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Newton’s Third Law of Motion: Symmetry in Forces

Problem 15

Problem 16

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Normal, Tension, and Other Example of Forces

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

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Problem Solving Strategies

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

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Further Applications of Newton’s Laws of Motion

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

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Further Applications of Newton’s Laws of Motion

Problem 52

Problem 53

Problem 54

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College Physics by Openstax Chapter 5 Problem 1

A physics major is cooking breakfast when he notices that the frictional force between his steel spatula and his Teflon frying pan is only 0.200 N. Knowing the coefficient of kinetic friction between the two materials, he quickly calculates the normal force. What is it?

Solution:

The formula for friction is

f=μk Nf=\mu _{k\:}N

When we solve for the normal force, NN, in terms of the other variables, we have

N=fμkN=\frac{f}{\mu _k}

The coefficient of kinetic friction is 0.04. Therefore, the normal force is

N=fμkN=0.200 newton0.04N=5.00 newton  (Answer)\begin{align*} N & =\frac{f}{\mu _k} \\ N & =\frac{0.200\:\text{newton}}{0.04} \\ N & =5.00\:\text{newton} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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