Tag Archives: average force

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*}
    Advertisements
    Advertisements