1. **State the problem:** A bullet of mass 0.5 g (which is 0.0005 kg) travels at 200 m/s and exits a tree at 100 m/s. We need to find the kinetic energy transferred to the tree. 2. **Formula used:** Kinetic energy (KE) is given by $$KE = \frac{1}{2} m v^2$$ where $m$ is mass and $v$ is velocity. 3. **Calculate initial kinetic energy:** $$KE_{initial} = \frac{1}{2} \times 0.0005 \times 200^2 = \frac{1}{2} \times 0.0005 \times 40000 = 10 \text{ J}$$ 4. **Calculate final kinetic energy:** $$KE_{final} = \frac{1}{2} \times 0.0005 \times 100^2 = \frac{1}{2} \times 0.0005 \times 10000 = 2.5 \text{ J}$$ 5. **Calculate kinetic energy transferred:** $$KE_{transferred} = KE_{initial} - KE_{final} = 10 - 2.5 = 7.5 \text{ J}$$ 6. **Answer:** The kinetic energy transferred from the bullet to the tree is 7.5 J. Therefore, the correct choice is d. 7.500 J.