Tag Archives: Newton’s Laws of Motion

College Physics by Openstax Chapter 4 Problem 4


Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of 50.0 N is exerted and the astronaut’s acceleration is measured to be 0.893 m/s2. (a) Calculate her mass. (b) By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. Discuss how this would affect the measurement of the astronaut’s acceleration. Propose a method in which recoil of the vehicle is avoided.


Solution:

We are given the following: \sum F = 50.0 \ \text{N}, and a=0.893 \ \text{m/s}^{2}.

Part A. We can solve for the mass, m by using Newton’s second law of motion.

\begin{align*}
\sum F & = ma \\
50.0 \ \text{N} & = m \left( 0.893 \ \text{m/s}^{2} \right) \\
m & = \frac{50.0 \ \text{N}}{0.893 \ \text{m/s}^{2}} \\
m & = 56.0 \ \text{kg}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B. The measured acceleration is equal to the sum of the accelerations of the astronauts and the ship. That is

a_{measured}=a_{astronaut}+a_{ship}

If a force acting on the astronaut came from something other than the spaceship, the spaceship would not undergo a recoil. \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)


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College Physics by Openstax Chapter 4 Problem 3


A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate its acceleration.


Solution:

From Newton’s Second Law of Motion, \sum F =ma. Substituting the given values, we have

\begin{align*}
\sum F & = ma \\
60.0 \ \text{N} & = \left( 4.50 \ \text{kg} \right) \ a \\
a & = \frac{60.0 \ \text{N}}{4.50 \ \text{kg}} \\
a & = 13.3 \ \text{m/s}^{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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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

Problem 22

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Centripetal Force

Problem 29

Problem 30

Problem 31

Problem 32

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Newton’s Universal Law of Gravitation

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

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Satellites and Kepler’s Laws: An Argument for Simplicity

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

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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

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 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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Grantham PHY220 Week 2 Problem 3

A bullet is fired from a gun at a shooting range. The bullet hits the ground after 0.32 seconds. How far did it travel horizontally and vertically in this time if it was fired at a velocity of 1100 m/s?

Solution:

Assuming that the gun was fired horizontally.

Consider the horizontal component of the motion.

v_{o_x}=1100\:m/s

\Delta x=V_{0_x}t=\left(1100\:m/s\right)\left(0.32\:s\right)=352\:m

Consider the vertical component.

v_{o_y}=0\:m/s

\Delta y=V_{0_y}t-\frac{1}{2}gt^2=0-\frac{1}{2}\left(9.80\:m/s^2\right)\left(0.32\:s\right)^2=-0.50

The negative sign of  \Delta y means that the bullet went downward.

Therefore, the bullet traveled 352 m horizontally and 0.50 m vertically downward in 0.32 seconds.

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=\mu _{k\:}N

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

N=\frac{f}{\mu _k}

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

\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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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 \ \text{kg} , and acceleration, a = 4.20 \ \text{m/s}^2.

The net force has a formula 

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

Substituting the given values, we have

\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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