A bullet of mass m = 20 g is fired at v_0 = 1000 m/s and…

A bullet of mass $m = 20\ \mathrm{g}$ is fired at $v_0 = 1000\ \mathrm{m/s}$ and hits the ground at $v = 200\ \mathrm{m/s}$. How much energy was spent overcoming air resistance? (Neglect the difference in height.)

Explanation of Method

Energy a bullet loses to air resistance.

Step 1: Kinetic energies

\[ E_0 = \tfrac12 \cdot 0.02 \cdot 1000^2 = 10\,000\ \mathrm{J}, \]

\[ E = \tfrac12 \cdot 0.02 \cdot 200^2 = 400\ \mathrm{J}. \]

Step 2: Energy lost

\[\Delta E = 9600\ \mathrm{J} = 9.6\ \mathrm{kJ}\]

$10\ \mathrm{kJ}$ $\text{at the gun}$ $\text{lost: } 9.6\ \mathrm{kJ}$ $\text{on impact}$ $0.4\ \mathrm{kJ}\text{ left}$

The bullet keeps only 4% of its kinetic energy; 96% goes into overcoming air resistance (heating the air and the bullet).

Result

\[\boxed{\,\Delta E = 9600\ \mathrm{J} = 9.6\ \mathrm{kJ}\,}\]

Remark

Check. The original result is correct (if the bullet lands at the height it was fired from; otherwise $mg\Delta h$ must be included).