Tag Archives: Force

College Physics by Openstax Chapter 8 Problem 8

A car moving at 10.0 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seat belt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70.0 kg.

Solution:

Newton’s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes. In symbols, Newton’s second law of motion is defined to be

Fnet=ΔpΔt\text{F}_\text{net} = \frac{\Delta \textbf{p}}{\Delta t}

Moreover, Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity.

p=mv\textbf{p} = m \textbf{v}

So, the Newton’s second law becomes

Fnet=mΔvΔt\text{F}_\text{net} = \frac{m\Delta\textbf{v}}{\Delta t}

Substituting the given values, we have

Fnet=mΔvΔtFnet=(70 kg)(10 m/s)0.26 sFnet=2692.3077 NFnet=2.69×103 N  (Answer)\begin{align*} \text{F}_\text{net} = & \frac{m \Delta\textbf{v}}{\Delta t} \\ \text{F}_\text{net} = & \frac{\left( 70\ \text{kg} \right)\left( 10\ \text{m/s} \right)}{0.26\ \text{s}} \\ \text{F}_\text{net} = & 2692.3077 \ \text{N} \\ \text{F}_\text{net} = & 2.69 \times 10^{3}\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics Chapter 8 Problem 7

A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average force exerted on a 0.0300-kg bullet to accelerate it to a speed of 600.0 m/s in a time of 2.00 ms (milliseconds)?

Solution:

Impulse, or change in momentum, equals the average net external force multiplied by the time this force acts:

Δp=FnetΔt\Delta \textbf{p}=F_{\text{net}} \Delta t

    The unknown in this problem is the average force. We can solve the formula for force in terms of the other quantities.

    Fnet=ΔpΔtF_{\text{net}} = \frac{\Delta \textbf{p}}{\Delta t}

    We are given the following values:

    Δp=mΔv=(0.0300 kg)(600.0 m/s)=18.00 kg⋅m/sΔt=2.00×10−3 s\begin{align*} \Delta \textbf{p} & = m \Delta v = \left( 0.0300\ \text{kg} \right)\left( 600.0\ \text{m/s} \right) = 18.00\ \text{kg}\cdot \text{m/s} \\ \Delta t& = 2.00 \times 10^{-3}\ \text{s} \end{align*}

    Substituting these given values, we can solve for the unknown.

    Fnet=ΔpΔtFnet=18.00 kg⋅m/s2.00×10−3 sFnet=9000 N  (Answer)\begin{align*} F_{\text{net}} & = \frac{\Delta \textbf{p}}{\Delta t} \\ F_{\text{net}} & = \frac{18.00\ \text{kg}\cdot \text{m/s}}{2.00 \times 10^{-3}\ \text{s}}\\ F_{\text{net}} & = 9000\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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    Statics 3.6 – Equilibrium of Truss Members Connected to a Gusset Plate | Hibbeler 14th Edition

    The gusset plate is subjected to the forces of three members. Determine the tension force in member C and its angle θ for equilibrium. The forces are concurrent at point O. Take F=8 kN.

    Engineering Mechanics: Statics 14th Edition by RC Hibbeler Problem 3-5 Equilibrium of Truss Members Connected in a Gusset Plate
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    Solution Guides to College Physics by Openstax Chapter 8 Banner

    Chapter 8: Linear Momentum and Collisions

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    Linear Momentum and Force

    Problem 1

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    Impulse

    Problem 7

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    Conservation of Momentum

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    Elastic Collisions in One Dimension

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    Inelastic Collisions in One Dimension

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    Collisions of Point Masses in Two Dimensions

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    Introduction to Rocket Propulsion

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    Electrical Transmission Towers Background for Force Vectors Statics of Rigid Bodies

    Chapter 2: Force Vectors

    Vector Addition of Forces

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    Addition of a System of Coplanar Forces

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    Cartesian Vectors | Addition of Cartesian Vectors

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    Force Vector Directed Along a Line

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    Dot Product

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