A 50 g bullet hits a target with a speed of 500 m/s and penetrates 25 cm without deviating from the initial
trajectory until it stops. Determine the magnitude of the average frictional force that resists penetration.
Problem data:
- Bullet bass: m = 50 g;
- Initial speed of the bullet: v0 = 500 m/s;
- Final speed of the bullet: v = 0;
- Distance that the bullet penetrates the target: d = 25 cm.
Problem diagram:
Solution
First, we convert the mass of the bullet given in grams (g) to kilograms (kg), and the penetration
distance given in centimeters (cm) to meters (m) used in the
International System of Units
(
SI)
\[
\begin{gather}
m=50\;\mathrm{\cancel g}\times\frac{1\;\mathrm{kg}}{1000\;\mathrm{\cancel{g}}}=0.05\;\mathrm{kg}\\[10pt]
d=25\;\mathrm{\cancel{cm}}\times\frac{1\;\mathrm m}{100\;\mathrm{\cancel{cm}}}=0.25\;\mathrm m
\end{gather}
\]
The work done by the force to stop the bullet is given by the
Work-Kinetic Energy Theorem, which is
equal to the change in the kinetic energy
\[
\begin{gather}
\bbox[#99CCFF,10px]
{W_{\small F}=\Delta K=K_f-K_i} \tag{I}
\end{gather}
\]
kinetic energy is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{K=\frac{mv^2}{2}} \tag{II}
\end{gather}
\]
substituting the equation (II) into equation (I) for the initial and final situations
\[
\begin{gather}
W_{\small F}=\frac{mv^2}{2}-\frac{mv_0^2}{2}\\[5pt]
W_{\small F}=\frac{(0.05\;\mathrm{kg})\times\left(0^2\right)}{2}-\frac{(0.05\;\mathrm{kg})\times\left(500\;\mathrm{\frac{m}{s}}\right)^2}{2}\\[5pt]
W_{\small F}=0-\frac{(0.05\;\mathrm{kg})\times\left(250000\;\mathrm{\frac{m^2}{s^2}}\right)}{2}\\[5pt]
W_{\small F}=-6250\;\mathrm J
\end{gather}
\]
The average force exerted by the target during deceleration will be
\[
\begin{gather}
\bbox[#99CCFF,10px]
{W_{\small F}=Fd}
\end{gather}
\]
\[
\begin{gather}
F=\frac{W_{\small F}}{d}\\[5pt]
F=\frac{-6250\;\mathrm J}{0.25\;\mathrm m}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{F=-25000\;\mathrm N}
\end{gather}
\]