In a ballistics test, a 28-g bullet pierces a sand bag that is 30 cm thick. If the initial bullet velocity was 55 m/s and it emerged from the sandbag moving at 18 m/s, what was the magnitude of the
Friction force (assuming it to be constant) that the bullet experienced while it traveled through the
Bag?

Question

Quizplus · Accepted Answer

The friction force can be found using the work-energy principle. The work done by the friction force is equal to the change in kinetic energy of the bullet. The initial kinetic energy (KE_i) is $ \frac{1}{2} m v_i^2 $ and the final kinetic energy (KE_f) is $ \frac{1}{2} m v_f^2 $, where $m$ is the mass of the bullet, $v_i$ is the initial velocity, and $v_f$ is the final velocity. The work done by the friction force is $ W = F_f d $, where $F_f$ is the friction force and $d$ is the distance traveled through the sandbag.Given:- $m = 28 \, \text{g} = 0.028 \, \text{kg}$ (mass must be in kg for SI units)- $v_i = 55 \, \text{m/s}$- $v_f = 18 \, \text{m/s}$- $d = 30 \, \text{cm} = 0.3 \, \text{m}$The change in kinetic energy ($ \Delta KE $) is $ KE_f - KE_i $:$$ \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 $$$$ \Delta KE = \frac{1}{2} \times 0.028 \times (18^2 - 55^2) $$$$ \Delta KE = 0.014 \times (324 - 3025) $$$$ \Delta KE = 0.014 \times (-2701) $$$$ \Delta KE = -37.814 \, \text{J} $$Since work done by the friction force is equal to the negative of the change in kinetic energy (because the friction force does work against the bullet's motion, reducing its kinetic energy),$$ W = - \Delta KE = 37.814 \, \text{J} $$The magnitude of the friction force is found by rearranging $ W = F_f d $ to solve for $ F_f $:$$ F_f = \frac{W}{d} = \frac{37.814}{0.3} \approx 126.05 \, \text{N} $$Rounding to the nearest option gives us 130 N.

In a Ballistics Test, a 28-G Bullet Pierces a Sand