Tag Archives: Force and Motion

Solution Guides to College Physics by Openstax Chapter 6 Banner

Chapter 6: Uniform Circular Motion and Gravitation

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Rotation Angle and Angular Velocity

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Centripetal Acceleration

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Centripetal Force

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Newton’s Universal Law of Gravitation

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Satellites and Kepler’s Laws: An Argument for Simplicity

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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

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Drag Forces

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Elasticity: Stress and Strain

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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

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Newton’s Third Law of Motion: Symmetry in Forces

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Normal, Tension, and Other Example of Forces

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Problem Solving Strategies

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Further Applications of Newton’s Laws of Motion

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Further Applications of Newton’s Laws of Motion

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Grantham PHY220 Week 2 Assignment Problem 8

If a car is traveling at 50 m/s and then stops over 300 meters (while sliding), what is the coefficient of kinetic friction between the tires of the car and the road?

SOLUTION:

Draw the free-body diagram of the car

week 2 problem 8

Consider the vertical direction

\sum F_y=ma_y

F_N-mg=0

F_N=mg

Consider the motion in the horizontal direction

Solve for the acceleration of the car.

v^2=\left(v_0\right)^2+2a_x\Delta x

a_x=\frac{v^2-\left(v_0\right)^2}{2\Delta x}=\frac{0-50^2}{2\left(300\right)}=-4.17\:m/s^2

Solve for the coefficient of kinetic friction

\sum F_x=ma_x

-F_{fr}=ma_x

-\mu _kF_N=ma_x

\mu _k=\frac{ma_x}{-F_N}=\frac{m\:\left(-4.17\right)}{-m\left(9.80\right)}=\frac{4.17}{9.80}

\mu _k=0.43

College Physics by Openstax Chapter 4 Problem 1

A 63.0-kg sprinter starts a race with an acceleration of 4.20 m/s2. What is the net external force on him?

Solution:

So, we are given mass, m=63.0 kg m = 63.0 \ \text{kg} , and acceleration, a=4.20 m/s2a = 4.20 \ \text{m/s}^2.

The net force has a formula 

F=ma\text{F}=\text{m}a

Substituting the given values, we have

F=(63.0 kg)(4.20 m/s2)F=265 kg⋅m/s2F=265 N  (Answer)\begin{align*} F & = \left( 63.0 \ \text{kg} \right)\left( 4.20 \ \text{m/s}^2 \right) \\ F & = 265 \ \text{kg}\cdot \text{m/s}^2 \\ F & = 265 \ \text{N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
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