4. (20 points) A bullet with a mass of $m=10$ g, flying horizontally at a speed of $v_{1}=400 \mathrm{m} / \mathrm{s}$, penetrates a massive board and exits with a speed of $v_{2}=100 \mathrm{m} / \mathrm{s}$. Find the magnitude of the work done on the bullet by the resistance force of the board.
Answer: 750 J

Solution. From the law of conservation of energy, it follows that

$$
A=\frac{m v_{1}^{2}}{2}-\frac{m v_{2}^{2}}{2}=\frac{0.01 \cdot 400^{2}}{2}-\frac{0.01 \cdot 100^{2}}{2}=750 \text { J. }
$$