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: ∑F=50.0 N\sum F = 50.0 \ \text{N}, and a=0.893 m/s2a=0.893 \ \text{m/s}^{2}.

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

∑F=ma50.0 N=m(0.893 m/s2)m=50.0 N0.893 m/s2m=56.0 kg  (Answer)\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

ameasured=aastronaut+ashipa_{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.   (Answer)\ \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, ∑F=ma \sum F =ma. Substituting the given values, we have

∑F=ma60.0 N=(4.50 kg) aa=60.0 N4.50 kga=13.3 m/s2  (Answer)\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

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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 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=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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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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